I love how you explain how to think in these challenging problems in stead of just doing everything correct right away, because many solutions can involve steps that most people wouldn't come up with themselves.
To get all the solution, when you have 2x^3 - (x-2)^3=0, you can factor out by using a^3 - b^3 = (a-b)(a^2 + ab + b^2) and we know that a = crt2*x and b = x-2 so we can substitute : (crt2*x-x+2)((crt2*x)^2 -crt2*x(x-2) + (x-2)^2) = 0 => [(crt2 - 1)x + 2][(crt2)^2*x^2 - crt2*x^2 + 2crt2*x + x^2 - 4x + 4] = 0 -> [(crt2 - 1)x + 2][(crt2^2 - crt2 + 1)*x^2 + (2crt2 - 4)*x + 4] = 0. And then, we get (crt2 - 1)x + 2 = 0 or (crt2^2 - crt2 +1)x^2 + (2crt2 - 4)x + 4 = 0 So for the first equation, we get x = - 2/(crt2-1) and for second, delta = 4*crt2^2 - 16*crt2 + 16 - 16crt2^2 + 16crt2 - 16 = - 12*crt2^2 so x1= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x2 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) So the answer are x1 = - 2/(crt2-1), x2= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x3 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) Crt = cuberoot and sqrt = squareroot
At 8:40 you take the cube root of both sides. Let w = (-1+sqrt(-3))/2. We can conclude cbrt(2) = x-2, but we could also conclude w*cbrt(2)=x-2, as well as w^2*cbrt(2) = x-2, and this gives the other two roots.
Since Euler own mistakes using √-3 notation, we avoid it by using i√3. Otherwise we get into trouble of arithmetic paradox and inconsistancies like : √-3×√-3=√9=3=-3=(√-3)^2
@@Igdrazil So the idea is to only put positive numbers in a square root sign. Seems good to me. Since the square root symbol means the positive root, that should help.
@@alnitaka Well it dépends in fact in the meaning one gives to this symbol √. First in precomplex algebra, one start with the quadratic function q(x)=x^2, which graphic representation is the well known conic : the parabolae.. Starting from there, the square root function is a priori the inverse function of the quadratic. And thus, its graphic representation is...also a parabolae...but quarter turned clockwise, or miror image of the quadratic, with respect to the first bissectrice y=x (in cartesian coordinate). This is the complete geometric story. BUT, we then have a first little problem..Since defined in this way the square root function, is no singled valued. But gives instead two image to any positive input. Zero also, can be seen to have two coïncidant images : zero ! So since mathematics is more a fresh clay pot than a granit stone, on can answer in at least two obvious ways to this situation. Either, one wants to stick to single valued functions for obvious commodity : single output to each inputs. And then there is two easy choics. Or define the square root of positive numbers as the positive hight of the square root total curve. Or as the negative. Traditionaly the positive one is chosen. BUT this DOESN'T mean that positive numbers have only one square root. They still ave two, thus noted √p and it's opposit - √p. And otherwise one can decide to play with multivalued functions and get by this door, closer to the garden.of Riemann surfaces and manifolds. Ok, that was a reminder to recall the maleability and open choices that stand in forging mathematic tools. So what about square root of negatives? From our former construction, it seems to make no sens at all. Nevertheless, one can forget the dressing and keep the spirit. And Indeed, what we want is to find, given a generalised number Z (négative or eaven a complex one living in the entire plane), the two opposit solutions to the equation z^2=Z. So to cut notations ambiguity, one can note R the square root bivalued function and keep √ symbol for the POSITIVE SQUARE ROOT OF POSITIVE REAL NUMBERS. But in practice, √ is often abusively used as a more or less ambiguous "square root" without saying what meaning it clearly takes. The context often helps. But unskilled students can mess around and make a m hell out of it by several nested confusions. The problem with such a notation as √z is that it tends to suggest that there is ONE number, whereas it gives two output. To overcome this multivalue problem Riemann invented surfaces and manifolds with several sheets, on which the previous multivalued function, becomes single valued on such manifolds. The square roots becomes single valued again on a two sheet Riemann surface. Leading to spin like behavior, of objects that come back to their initial state, not after one full turn, but after two. Like a book standing flat on the open hand. Thus multivalued functions are not "problematic", but on the contrary they bring lots of new marvelous insight and jucy structures. But one has to be more carefull with them. And for instance, for obvious reasons of multioutput, the old MORPHISME structure of √ : √a×√b = √(ab), is lost ! Unless one puts itself on the Riemann surface where no ambiguity remains. And as a practical tool, we think of √ as ^(1/2), applied on a complex number r.exp(ia) which phase a is defined modulo 2π. By this method and notation, all roots are taken into acount, thanks to the 2π periodicity, that is cut in half by the power 1/2. By this mean all confusions are killed. So with this in mind, if -1 is understood as the complex numbers exp(iπ) AND any cyclic other candidate, like exp(3iπ), then taking the square root as a 1/2 power on those two, leads correctly to the two solutions :exp(iπ\2)=i and exp(3π/2)=-i. Similar subtilities arises with higher n'th roots or the logarithm of a complex number...
8:18 There are two more solutions. 2x^3-(x-2)3=[2^(1/3)x-(x-2)][2^(2/3)x^2+2^(1/3)x(x-2)+(x-2)^2]=0 First term is real solution (SyberMath solution). Second term giver two complex conjugate solutions.
One can divide the equation by 8 and let y = -x/2. So the equation becomes: -y^3 + 3y^2 + 3y + 1 = 0. 2y^3 = y^3 + 3y^2 + 3y + 1 = (y + 1)^3. 2 = (1 + 1/y)^3. Taking cubic root, 1 + 1/y = r, where r = cubicRoot(2). There are 3 cubic roots of 2. r represents any of them. y = 1 / [r - 1] = [r^2 + r + 1 ]. As y = -x/2, x = -2 * [r^2 + r + 1 ]. ...
Nice trick! It was the third one I tried. Finding the complex roots isn't too hard, but writing them on the screen would make everyone's eyes glaze over.
Sir, you have showed only the real solution and rationalised the denominator but there are also two other complex roots(also because by definition, a cubic equation has three roots) consisting of omega(cube root of 1), thank you for this solution....
@@alessandrofabi9252 One of the two (not trivialy one) cube root of unity, is noted usualy w or j. It's the point in the cartesian plane (-1/2;√3/2), or in complex algebraic form (-1+i√3)/2, or in exponential form exp(2iπ/3), or in complex trigonometric form cos(2π/3)+i.sin(2π/3). Where i stands for the similar square root of unity, being the point (0,1) in the cartesian plane, or the 2×2 Real Matrix of the form [0 1 / -1 0], which Indeed squares to "-1" which is the negative identity Matrix [-1 0 / 0 -1]. Such distinct of one, cube root of unity j or w satisfies the cubic equation : j^3=1, and hence also j^2+j+1=0, since j^3-1 factors out as (j-1)(j^2+j+1). In this form j^2+j+1=0 one can observe the three cube roots of unity (1 included), ADDING TO ZERO. That says in an other way that j and j^2 are complex conjugate, of unit lenght, with real part equal to -1/2 for both. Only their imaginary part are opposit : √3/2 and -√3/2. They form a multiplicative subgroup of the circle S1, whith obvious inverses 1/j=j^2, and 1/j^2=j, since j^3=1. Its apparent simplicity can be deceptive since it took for instance all Gauss genious to solve the 17 gone construction problem with rule and compas. Even the pentagone construction is tricky but lovely.
x³+6x²-12x+8=0 The coefficients hint at a perfect cube, dividing by the binomial coefficients ( 1 3 3 1) we get 1 2 -4 8 neatly powers of two (signs are a mess) then, as per video for 3 steps (x+2)³=x³+6x²+12x+8 (x-2)³=x³-6x²+12x-8 so 2x³-(x-2)³=x³+6x²-12x+8=0 2x³=(x-2)³ or ( (∛2.x)/(x-2) )³=1 (NB ∛ is the cube root - it doesn't look very readable on screen) Euler's formula: e^iθ = cosθ + i sinθ 1=e^(2inπ) has (3rd) roots at n=0,1,2 i.e. e^0, e^(2iπ/3), e^(4iπ/3) so 1, -(1/2)±(i√3/2) for a solution s (one of the above) (∛2.x)/(x-2)=s x(∛2-s)=-2s x=2s/(s-∛2) s=1 => 2/(1-∛2) × (1+∛2+(∛2)²)/(1+∛2+(∛2)²) x=2(1+∛2+(∛2)²)/(1-2) x=-2-2^(4/3)-2^(5/3) s=-(1/2)±(i√3/2) x=2*(-(1/2)±(i√3/2))/(-(1/2)±(i√3/2)-∛2) =2*(-(1/2)±(i√3/2))/(-(1/2+∛2)±(i√3/2)) after a whole mess of rearranging, the other solutions are x = -2+∛2(1+i√3)+(∛2)²(1-i√3) x = -2+∛2(1-i√3)+(∛2)²(1+i√3)
Hands-down the best mathematician on TH-cam he and his problems are unmatched and every single problem applies so many different concepts. His explanation is always on point.
Following my standard cubic method was less grungy than usual with this equation. I wound up with that answer, but I tried in vain to simplify it. I think that it took about as long as your method.
All these different seemingly random equations with all these different tricks.I think this is crazy in one sense at least.Yes,algebraic skill is developed and that is laudable,but it makes me feel that something is amiss with all this stuff.Structural relations between different objects to find equation solving universality I crave.I pray to the mathematical gods to let me get to the bottom of all this.
@@SyberMath So far I got that this expression equals (x+2)^3 -12x but NOW WHAT..surely this is what most very intelligent people wkuld do..i havent watched your solution yet but what then do you have (x +2)^3= 24x and take the cube root of both sides?? What else i don't see why wohld multiply the equation by something because there's no pattern tnere or to dividie by anything..hope you can resoo d..
Sub y = x + 2 to get y³ - 24y + 48 = 0 and Cardano it. I see the trick you were trying to show, but the messing around with the irrational denominator undid all the good work.
Hard to find one book but I'm compiling a list of good books and will share them in the community tab soon. Feel free to remind me if I forget! th-cam.com/channels/W4czokv40JYR-w7u6aXZ3g.htmlcommunity
Tricks work if somebody gives you a cubic where they work. Chang one coefficient or a sign and you may have to throw in the towel. Give me something that works ALL the time, without tricks.
There is several. First depress your original cubic by getting rid of the quadratic termes by a horizontal translation.. Then you can use the historic method giving yourself freedom by looking at a solution x of the form u+v. By substitution in the depressed cubic, you impose a permissive condition on u and v in order to get to two non linear but simple to solve equations for (uv)^3=p and u^3+v^3=s. Noting U=u^3 and V=v^3 you get the well known quadratic solutions U and V, satisfying U+V =s and UV=p, and hence solutions of the quadratic X^2-sX+p=0. This gives you your three complex solutions. But there is much nicer way, straight forward, using an algebraic identity...
I love how you explain how to think in these challenging problems in stead of just doing everything correct right away, because many solutions can involve steps that most people wouldn't come up with themselves.
Thank you!
That's depressing why do you say most ppl woukdnt come up with themselves..are most ppl not very smart then?
Coming up with something like this in your own is something only extremely smart people are likely to figure out.
To get all the solution, when you have 2x^3 - (x-2)^3=0, you can factor out by using a^3 - b^3 = (a-b)(a^2 + ab + b^2) and we know that a = crt2*x and b = x-2 so we can substitute : (crt2*x-x+2)((crt2*x)^2 -crt2*x(x-2) + (x-2)^2) = 0 => [(crt2 - 1)x + 2][(crt2)^2*x^2 - crt2*x^2 + 2crt2*x + x^2 - 4x + 4] = 0 -> [(crt2 - 1)x + 2][(crt2^2 - crt2 + 1)*x^2 + (2crt2 - 4)*x + 4] = 0. And then, we get (crt2 - 1)x + 2 = 0 or (crt2^2 - crt2 +1)x^2 + (2crt2 - 4)x + 4 = 0
So for the first equation, we get x = - 2/(crt2-1) and for second, delta = 4*crt2^2 - 16*crt2 + 16 - 16crt2^2 + 16crt2 - 16 = - 12*crt2^2 so x1= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x2 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
So the answer are x1 = - 2/(crt2-1), x2= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x3 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
Crt = cuberoot and sqrt = squareroot
At 8:40 you take the cube root of both sides. Let w = (-1+sqrt(-3))/2. We can conclude cbrt(2) = x-2, but we could also conclude w*cbrt(2)=x-2, as well as w^2*cbrt(2) = x-2, and this gives the other two roots.
Nice!
Since Euler own mistakes using √-3 notation, we avoid it by using i√3. Otherwise we get into trouble of arithmetic paradox and inconsistancies like : √-3×√-3=√9=3=-3=(√-3)^2
@@Igdrazil So the idea is to only put positive numbers in a square root sign. Seems good to me. Since the square root symbol means the positive root, that should help.
@@alnitaka Well it dépends in fact in the meaning one gives to this symbol √. First in precomplex algebra, one start with the quadratic function q(x)=x^2, which graphic representation is the well known conic : the parabolae..
Starting from there, the square root function is a priori the inverse function of the quadratic. And thus, its graphic representation is...also a parabolae...but quarter turned clockwise, or miror image of the quadratic, with respect to the first bissectrice y=x (in cartesian coordinate). This is the complete geometric story.
BUT, we then have a first little problem..Since defined in this way the square root function, is no singled valued. But gives instead two image to any positive input. Zero also, can be seen to have two coïncidant images : zero !
So since mathematics is more a fresh clay pot than a granit stone, on can answer in at least two obvious ways to this situation.
Either, one wants to stick to single valued functions for obvious commodity : single output to each inputs. And then there is two easy choics. Or define the square root of positive numbers as the positive hight of the square root total curve. Or as the negative. Traditionaly the positive one is chosen.
BUT this DOESN'T mean that positive numbers have only one square root. They still ave two, thus noted √p and it's opposit - √p.
And otherwise one can decide to play with multivalued functions and get by this door, closer to the garden.of Riemann surfaces and manifolds.
Ok, that was a reminder to recall the maleability and open choices that stand in forging mathematic tools.
So what about square root of negatives? From our former construction, it seems to make no sens at all. Nevertheless, one can forget the dressing and keep the spirit.
And Indeed, what we want is to find, given a generalised number Z (négative or eaven a complex one living in the entire plane), the two opposit solutions to the equation z^2=Z.
So to cut notations ambiguity, one can note R the square root bivalued function and keep √ symbol for the POSITIVE SQUARE ROOT OF POSITIVE REAL NUMBERS.
But in practice, √ is often abusively used as a more or less ambiguous "square root" without saying what meaning it clearly takes. The context often helps. But unskilled students can mess around and make a m
hell out of it by several nested confusions.
The problem with such a notation as √z is that it tends to suggest that there is ONE number, whereas it gives two output.
To overcome this multivalue problem Riemann invented surfaces and manifolds with several sheets, on which the previous multivalued function, becomes single valued on such manifolds.
The square roots becomes single valued again on a two sheet Riemann surface. Leading to spin like behavior, of objects that come back to their initial state, not after one full turn, but after two. Like a book standing flat on the open hand.
Thus multivalued functions are not "problematic", but on the contrary they bring lots of new marvelous insight and jucy structures.
But one has to be more carefull with them. And for instance, for obvious reasons of multioutput, the old MORPHISME structure of √ :
√a×√b = √(ab), is lost ! Unless one puts itself on the Riemann surface where no ambiguity remains.
And as a practical tool, we think of √ as ^(1/2), applied on a complex number r.exp(ia) which phase a is defined modulo 2π. By this method and notation, all roots are taken into acount, thanks to the 2π periodicity, that is cut in half by the power 1/2.
By this mean all confusions are killed.
So with this in mind, if -1 is understood as the complex numbers exp(iπ) AND any cyclic other candidate, like exp(3iπ), then taking the square root as a 1/2 power on those two, leads correctly to the two solutions :exp(iπ\2)=i and exp(3π/2)=-i.
Similar subtilities arises with higher n'th roots or the logarithm of a complex number...
8:18 There are two more solutions. 2x^3-(x-2)3=[2^(1/3)x-(x-2)][2^(2/3)x^2+2^(1/3)x(x-2)+(x-2)^2]=0 First term is real solution (SyberMath solution). Second term giver two complex conjugate solutions.
I usually ignore the complex solutions but you're right!
This is smart!
Thanks :)
One can divide the equation by 8 and let y = -x/2. So the equation becomes: -y^3 + 3y^2 + 3y + 1 = 0. 2y^3 = y^3 + 3y^2 + 3y + 1 = (y + 1)^3.
2 = (1 + 1/y)^3. Taking cubic root, 1 + 1/y = r, where r = cubicRoot(2). There are 3 cubic roots of 2. r represents any of them.
y = 1 / [r - 1] = [r^2 + r + 1 ]. As y = -x/2, x = -2 * [r^2 + r + 1 ].
...
Teaching people the thought process to search for a solution. Very good!
Nice trick! It was the third one I tried. Finding the complex roots isn't too hard, but writing them on the screen would make everyone's eyes glaze over.
Sir, you have showed only the real solution and rationalised the denominator but there are also two other complex roots(also because by definition, a cubic equation has three roots) consisting of omega(cube root of 1), thank you for this solution....
No problem. Yes there are complex solutions!
Just to understand: what’s omega?
@@alessandrofabi9252 One of the two (not trivialy one) cube root of unity, is noted usualy w or j. It's the point in the cartesian plane (-1/2;√3/2), or in complex algebraic form (-1+i√3)/2, or in exponential form exp(2iπ/3), or in complex trigonometric form cos(2π/3)+i.sin(2π/3).
Where i stands for the similar square root of unity, being the point (0,1) in the cartesian plane, or the 2×2 Real Matrix of the form [0 1 / -1 0], which Indeed squares to "-1" which is the negative identity Matrix [-1 0 / 0 -1].
Such distinct of one, cube root of unity j or w satisfies the cubic equation : j^3=1, and hence also j^2+j+1=0, since j^3-1 factors out as (j-1)(j^2+j+1). In this form j^2+j+1=0 one can observe the three cube roots of unity (1 included), ADDING TO ZERO. That says in an other way that j and j^2 are complex conjugate, of unit lenght, with real part equal to -1/2 for both. Only their imaginary part are opposit : √3/2 and -√3/2.
They form a multiplicative subgroup of the circle S1, whith obvious inverses 1/j=j^2, and 1/j^2=j, since j^3=1.
Its apparent simplicity can be deceptive since it took for instance all Gauss genious to solve the 17 gone construction problem with rule and compas. Even the pentagone construction is tricky but lovely.
x³+6x²-12x+8=0
The coefficients hint at a perfect cube, dividing by the binomial coefficients ( 1 3 3 1)
we get 1 2 -4 8 neatly powers of two (signs are a mess) then, as per video for 3 steps
(x+2)³=x³+6x²+12x+8
(x-2)³=x³-6x²+12x-8
so 2x³-(x-2)³=x³+6x²-12x+8=0
2x³=(x-2)³
or ( (∛2.x)/(x-2) )³=1 (NB ∛ is the cube root - it doesn't look very readable on screen)
Euler's formula: e^iθ = cosθ + i sinθ
1=e^(2inπ) has (3rd) roots at n=0,1,2 i.e. e^0, e^(2iπ/3), e^(4iπ/3) so 1, -(1/2)±(i√3/2)
for a solution s (one of the above)
(∛2.x)/(x-2)=s
x(∛2-s)=-2s
x=2s/(s-∛2)
s=1 => 2/(1-∛2) × (1+∛2+(∛2)²)/(1+∛2+(∛2)²)
x=2(1+∛2+(∛2)²)/(1-2)
x=-2-2^(4/3)-2^(5/3)
s=-(1/2)±(i√3/2)
x=2*(-(1/2)±(i√3/2))/(-(1/2)±(i√3/2)-∛2)
=2*(-(1/2)±(i√3/2))/(-(1/2+∛2)±(i√3/2))
after a whole mess of rearranging, the other solutions are
x = -2+∛2(1+i√3)+(∛2)²(1-i√3)
x = -2+∛2(1-i√3)+(∛2)²(1+i√3)
Solving this equation using lagrange’s resolvent and cardano’s formula in another video would be great sir!
Hands-down the best mathematician on TH-cam he and his problems are unmatched and every single problem applies so many different concepts. His explanation is always on point.
He's actually not a mathematician - he just enjoys solving math problems. However, he would be very flattered!
How about this method of using (x+y)^3 pattern ?
(x+2)^3 = x^3 + 3(x)(2)(x+2) + 2^3 = 24x = 24(x+2) - 48
Let y = x+2 and m = -8 and n = -24. Then, y^3 + 3my = 2n.
So, cubic-discriminant D = n^2 + m^3 = 576 - 512 = 64 = 8^2. Therefore x+2 = y = ∛(n + √D) + ∛(n - √D) = ∛(-24 + √64) + ∛(-24 - √64) = - ∛32 - ∛16
Hence, *x = -2 - ∛32 - ∛16*
Beautiful way of solving the problem,enriching my knowledge.🙏🙏
It's my pleasure!
Following my standard cubic method was less grungy than usual with this equation. I wound up with that answer, but I tried in vain to simplify it. I think that it took about as long as your method.
...nice job explaining the real number answer. Very smooth!
Thanks!
All these different seemingly random equations with all these different tricks.I think this is crazy in one sense at least.Yes,algebraic skill is developed and that is laudable,but it makes me feel that something is amiss with all this stuff.Structural relations between different objects to find equation solving universality I crave.I pray to the mathematical gods to let me get to the bottom of all this.
Great solution. Thank you. I learned one more.
Great to hear!
@@SyberMath So far I got that this expression equals (x+2)^3 -12x but NOW WHAT..surely this is what most very intelligent people wkuld do..i havent watched your solution yet but what then do you have (x +2)^3= 24x and take the cube root of both sides?? What else i don't see why wohld multiply the equation by something because there's no pattern tnere or to dividie by anything..hope you can resoo d..
Magnificent mathematical innovation!
Very nice. If you don’t have a calculator handy you can approximate the root using recursion and the system y=24x, y=(x+2)^3 with x[0] less than zero.
More succinctly, just use the recursive relationship x[i+1]=-(1/24)(-x[i]+2)^3 with x[0]
5:45 This line seems like my native language dialogue 😂
That’s is so amazing trick and clearly how to do it ,great explaining!
why the crossing out. the factor of (x-2) is clear. so (x-2)(x^2-4) +6x(x-2)=0
Alt.
(2-x)^3+2x^3=0
B/c you can flip the 2 to the front part of the equation. It's a bit more cleaner to.
How would you know to arrange it that way?
David Flanders Let 2=y, and notice how in a binomial expansion the signs alternate. Variables are just #'s.
I am bangladeshi .but your class is very nice so I see and flow your lecture
Many many thanks. Greetings from the United States! 💖
Could you solve this equation using cardano’s formula sir? Thank you a lot sir
thx again dear teacher ..
You're welcome 😊
Nice trick!
Sub y = x + 2 to get y³ - 24y + 48 = 0 and Cardano it. I see the trick you were trying to show, but the messing around with the irrational denominator undid all the good work.
Nice!
A great one!
Can you do more cubic equation with all type of solution please?
Awesome!
That's ONE or the solutions. What about the other 2?
You are amazing !
Moço, você é maravilhoso. Conhece demais.(Brasil). (Young man, you are wonderful. You know too much.)
Great!
Can you suggest a book which has problems in algebra of this level
Hard to find one book but I'm compiling a list of good books and will share them in the community tab soon. Feel free to remind me if I forget!
th-cam.com/channels/W4czokv40JYR-w7u6aXZ3g.htmlcommunity
Nice!
So beautiful! 😇👏👏👏
Beatifull amazing maths is beaty
Especial thanks to thinking loudly for solving this. I like this method becsuse of being normal .thanks again.
Most welcome 😊
Simply multiply whole equation with (-1). Further the given equation must change in to (a-b) ^3 form
Tjat doesn't do enough..that just gives you something that equals-(x -2)^3 but then what? I don't see what you can do
Unbelievable
You hsce x minus a multiply x minus b multiply x minus c is equal to that cubic equstiion too
Thats great
-2 is also a real solution how to find that
smarter than smartest...
i got solution
thanks ❤️
Thank you for the kind words! 💖
I lobe this sooooo much :D
Thanks!
So neat.
Thanks!
Why didn’t you just find one factor then use long division?? Would have been quicker than all this mucking about.
Very good
Thank you
3X+12X-12X+8=0
You are best
Thank you! 😍
Tricks work if somebody gives you a cubic where they work. Chang one coefficient or a sign and you may have to throw in the towel. Give me something that works ALL the time, without tricks.
There is several. First depress your original cubic by getting rid of the quadratic termes by a horizontal translation.. Then you can use the historic method giving yourself freedom by looking at a solution x of the form u+v. By substitution in the depressed cubic, you impose a permissive condition on u and v in order to get to two non linear but simple to solve equations for (uv)^3=p and u^3+v^3=s. Noting U=u^3 and V=v^3 you get the well known quadratic solutions U and V, satisfying U+V =s and UV=p, and hence solutions of the quadratic X^2-sX+p=0.
This gives you your three complex solutions.
But there is much nicer way, straight forward, using an algebraic identity...
Please don’t write at the bottom of the screen. We can’t see under the subtitles.
I keep forgetting 😁
awesome
Thank you, Antoine! 🤩
it was lengthy problem and not tricky
This is wrong, x should have 3 answers. Not one answer. x=2, x=2(sqrt(3)-2) , x=-2(sqrt(3)+2)
All the cubic equations are required to have three answers?
-7.69
Mathemagic👍
Hehe! Thanks! 😁
In exam you do not have so much time to try all that
Right but you have time if you're not taking a test
👍
Abrakadabra....🤣
😂 Ehehe! This is like magic, isn't it?
👍👍👍
apenii😂xpahamla
For God’s sake remove the subtitles especially that you are talking English!
You can turn off the subtitles
I got x=-2.
How?
-(2*2*2)+6*-1(2*2)-12(-2)+8=0. That's how I got x=-2. Check it out.
It's -8-24+24+8=0
@@carloshuertas4734 doesn't work
@@carloshuertas4734 you multiplied wrong on the x² term