It's damm bulender false😳, becoz, you comparing LHS and RHS, after Breaking RHS in to sum of 2 number square, but, you only do for integer, there are infinite number for which 36 can be broken in two number's sum, but you took only the case of integer! So srry, it's not complete
It's damm bulender false😳, becoz, you comparing LHS and RHS, after Breaking RHS in to sum of 2 number square, but, you only do for integer, there are infinite number for which 36 can be broken in two number's sum, but you took only the case of integer! So srry, it's not complete
I did it in 30 seconds. As I saw powers 9 and 6, immediately a thought came in my mind that I should take a^3=t. So it will be t^3 + t^2=36 t^2 (t+1)= (3)^2 (3+1) So t=3, a^3=3. a=(3)^1/3
or u could just do it by taking a to the power 6 common and u will see that in the lhs its a multiplication of a factor of a and a non factor of a so u could do same for 36 and divide it into its prime factors 2 and 3 and in rhs u would get 9 * 4 and by equating terms u would get answer pretty quickly
{3^(1/3)}is the solution. Consider a³ = t. and then simple cubic equation will be formed. By hit and trial t= 3 satisfies this equation. and from this , value of a will be obtained. Edit 1 - I didn't know that sir used the same method , but i don't copy him , i did it myself with watching the solution by sir.
a⁹+a⁶= 36 a⁶(a³+1)=36 a⁶(a³+1)= (2×3)² a⁶(a³+1)=2² × 3² On comparing both sides, a=³√2 or a= ³√3 Put both values in eq.... And only (a= ³√3) will satisfy....
Sir, as we know that quadratic equation has always two solutions and cubic equation has three solutions, we simply say that the number of solutions satisfying an equation is directly proportional to the degree of the equation. This equation is an equation of degree 9 if a=(3)^1/3 this only one value of , so how can find out the other 8 values of a?
I solved this question by simple method 😂 I split the value 36 into 27 + 9 Now equate this value with a⁹+a⁶,as a⁹ + a⁶ = 27 + 9 a⁹ + a⁶ = 3³ + 3² By comparing we get a⁹ = 3³. or a⁶ = 3² a = 3⅓
Is equation ka answer 18 bhi aa sakta hai agr a common nikal le aur 1 ki power 9 toh 1 hi hoga aur 1 ki power 6 toh 1 hi hoga fir 1 plus 1 2 hi hoga 2 36 ke neeche chala jayega aur a ki value 18 aa jayega
(a³+1)a⁶=36 , if a³ is even then (a³ +1) is odd and if a³ is odd( a³+1) is even , so basically we have to break 36 is product if even and odd , very few cases are there so just write them and check .
We can solve it like a9 + a6 = 36. a9 + a6 = 3² x 2² a6(a³+1) = 3²x2² If a³ + 1 = 3² ; a = 2.. And if a³ + 1 =2² ; a = 3⅓.. On substituing both a = 3⅓ satisfues hence it is the solution😁
Sir isko aise bhi to kar sakte hai.. a⁶+a⁹=36 a⁶(a³+1)=36 say a³=x , then x²(x+1)=36 x³+x²=36 Clear dikh rha hai ki 27+9 satisfy kar rha hai therefore x³+x²=27+9 x³+x²=3³+3² Therefore, x=3 So , a³=3 Hence , a=3⅓ 💯
a⁶(a³+1)=36 so product of 2 numbers =36 hona chahiye aur aaisa and no 1 should be greater than no 2 Cases: 18,2 12,3 9,4 6,6(nhi hoga) Individually solve krdiya 3⅓ agya 😅😅3 min me done.
Solved in a diff. manner t²(t+1)=36 t²(t+1)=9(4) t²(t+1)=3²(3+1) Comparing both sides t=3 a=3⅓ After seeing sir's solution Realised that my solution is not so good...
Well this soloution could easily be solved in the following way... (good observation skills are needed)\ a^9 + a^6 = 36 a^6(a^3 + 1)= 9(3+1) Now anyone can fig. it out that a^3=3 ......
Sir please number theory pe lecture series laayiye. Please sir.please sir.Indian Statistical institute ke admission test ke liye seekhni hai number theory.please sir,please sir,please sir ..............∞
I don't know how it came to my mind after considering a³ = x ( substitution) And then x²(x+1)=36 Then x²(x+1) -36 =0 x³+x²-36=0 x³+ (x-6)(x+6) =0 Adding and subtracting 216 , x³ + 216+(x-6)(x+6)=216 How did I thought?? My attempt was to convert given question into two simple brackets in product. Now applying a³+b³ =a+b(a²+b²-ab) (x+6)(x²+36-6x) +(x-6)(x+6)=216 (x+6)(x²+36-6x +x-6) =216 Now interesting part begins!! I randomly putted x=6 in above equation and answer came as 432. Which is twice of 216.. So I thought to try it with x=3 and actually it satisfied. This is just by God grace. Random thought it was . I don't know how it came to my mind. We have substituted a³=x and therefore a³=3 And then a³-3=0 And after applying a³-b³= (a-b)(a²+b²-ab) We get (a-3⅓)(a² +3⅔+3⅓a)= 0 Therefore either a=3⅓ or a² +3⅔+3⅓a =0 Clearly we can see that D of Quadratic formed is less than 0 and a( ax²+bx+c) wala is 1 which is greater than 0 and therefore given Quadratic is always greater than 0. Therefore we get a=3⅓....
Uhm, let me tell u how I solved this problem at first glance:- a^9+a^6=36 1. Take a^6 common a^6(a^3+1)=36 2. Factorize 36 a^6(a^3+1)=2^2.3^2 3. Now just do some hit and trial by assuming either 2^2 or 3^2 to be a^6 You'll get a=3^1/3 or a=2^1/3 4.Insert the values of a in the original equation and check whether it equates to 36 or not 5. After checking u will get the answer as a=3^1/3.
![[TH] FS vs BME - VCT Ascension Pacific - Lower Bracket Final](http://i.ytimg.com/vi/wnygqsR2Fig/mqdefault.jpg)
Nice one. You may have used the SYNTHETIC DIVISION for dividing by the factor (t-3)- this saves time- and that's critical in an Olympiad.
Could u pls share a video regarding synthetic division.
Will synthetic division give the other factor?
@@uttkarsh7823 yeah but in this case all other 8 roots are imaginary
I solved this in 2min but not by your method 😊
It's so easy :
a⁹+a⁶=36
a⁹+a⁶= 3³+3² ( 36 can be written as
Same I did
But brother if next factor will be factorise than it's have 2 extra factor
It's damm bulender false😳, becoz, you comparing LHS and RHS, after Breaking RHS in to sum of 2 number square, but, you only do for integer, there are infinite number for which 36 can be broken in two number's sum, but you took only the case of integer! So srry, it's not complete
It's damm bulender false😳, becoz, you comparing LHS and RHS, after Breaking RHS in to sum of 2 number square, but, you only do for integer, there are infinite number for which 36 can be broken in two number's sum, but you took only the case of integer! So srry, it's not complete
Same I even tried to approximate..1.45
I did it in 30 seconds. As I saw powers 9 and 6, immediately a thought came in my mind that I should take a^3=t. So it will be t^3 + t^2=36
t^2 (t+1)= (3)^2 (3+1)
So t=3,
a^3=3.
a=(3)^1/3
But aapko ye kaise pta chala ki baki ke 8 root real nahi h
Qrdtc dekh ke 😂
Same way
For quick calc.
or u could just do it by taking a to the power 6 common and u will see that in the lhs its a multiplication of a factor of a and a non factor of a so u could do same for 36 and divide it into its prime factors 2 and 3 and in rhs u would get 9 * 4 and by equating terms u would get answer pretty quickly
I did it in 30 seconds too
Sir when u get the value of t = 3 then u could put that in a^3=t , then u get a^3=3 and than cube root both side u would directly get 3^⅓
{3^(1/3)}is the solution. Consider a³ = t. and then simple cubic equation will be formed. By hit and trial t= 3 satisfies this equation. and from this , value of a will be obtained.
Edit 1 - I didn't know that sir used the same method , but i don't copy him , i did it myself with watching the solution by sir.
It is also solved by heat and trial method when we expressed 36 in the form of 3 square and 2 square and then compare
a⁹+a⁶= 36
a⁶(a³+1)=36
a⁶(a³+1)= (2×3)²
a⁶(a³+1)=2² × 3²
On comparing both sides,
a=³√2 or a= ³√3
Put both values in eq....
And only (a= ³√3) will satisfy....
Sir, as we know that quadratic equation has always two solutions and cubic equation has three solutions, we simply say that the number of solutions satisfying an equation is directly proportional to the degree of the equation. This equation is an equation of degree 9 if a=(3)^1/3 this only one value of , so how can find out the other 8 values of a?
Other 8 values are imaginary.
you can do it by cube roots of unity
complex numbers ka part hai ye
I solved this question by simple method 😂
I split the value 36 into 27 + 9
Now equate this value with a⁹+a⁶,as
a⁹ + a⁶ = 27 + 9
a⁹ + a⁶ = 3³ + 3²
By comparing we get
a⁹ = 3³. or a⁶ = 3²
a = 3⅓
😂😂😂
real genius
more videos like this sir appreciate it a lot
I just saw the first step in which you substituted a^3 = t and there onward I did it easily with class 9th principles
Kuch bada ukhad nhi liya bhai
@@voterofNOTA ☺☺
I solve this problem in my first attemp by simple method that I learn from my 10th class
Is equation ka answer 18 bhi aa sakta hai agr a common nikal le aur 1 ki power 9 toh 1 hi hoga aur 1 ki power 6 toh 1 hi hoga fir 1 plus 1 2 hi hoga 2 36 ke neeche chala jayega aur a ki value 18 aa jayega
Very interesting presentation. Thanks !
Bhannat explanation sir ji 😎🔥🔥
pehli baar sir ka koi question solve hua
(a³+1)a⁶=36 , if a³ is even then (a³ +1) is odd and if a³ is odd( a³+1) is even , so basically we have to break 36 is product if even and odd , very few cases are there so just write them and check .
class 10th icse emainder nd factor theorem ka sawaal hai ye toh
turant bn gyaa😃
We can solve it like a9 + a6 = 36.
a9 + a6 = 3² x 2²
a6(a³+1) = 3²x2²
If a³ + 1 = 3² ; a = 2..
And if a³ + 1 =2² ; a = 3⅓..
On substituing both a = 3⅓ satisfues hence it is the solution😁
Sir isko aise bhi to kar sakte hai..
a⁶+a⁹=36
a⁶(a³+1)=36
say a³=x , then
x²(x+1)=36
x³+x²=36
Clear dikh rha hai ki 27+9 satisfy kar rha hai therefore
x³+x²=27+9
x³+x²=3³+3²
Therefore, x=3
So , a³=3
Hence , a=3⅓ 💯
😌😌feeling when you get the answer by yourself.
3^1/3, 3^1/3 omega, 3^1/3 omega square is the only correct answer
a⁶(a³+1)=36 so product of 2 numbers =36 hona chahiye aur aaisa and no 1 should be greater than no 2
Cases:
18,2
12,3
9,4
6,6(nhi hoga)
Individually solve krdiya 3⅓ agya
😅😅3 min me done.
I solved it by first dividing 36 to 4 and 9 then writing 4 as 3+1 then taking. a^6 common in LHS then equation from which we get a = 3^1/3
3^(1/3)
Solved in a diff. manner
t²(t+1)=36
t²(t+1)=9(4)
t²(t+1)=3²(3+1)
Comparing both sides
t=3
a=3⅓
After seeing sir's solution
Realised that my solution is not so good...
Sir pls bring more general questions for jee advanced
Mains hogyii?
@@sagarbambal5954 currently in 12th
Preparing for JEE 2024
This is too easy for someone of your caliber 😂
Sir i had also done the same method but i got stuck when i didn't know how to solve cubic questions
Well this soloution could easily be solved in the following way... (good observation skills are needed)\
a^9 + a^6 = 36
a^6(a^3 + 1)= 9(3+1)
Now anyone can fig. it out that a^3=3 ......
As Olympiads contain mcqs, trial and error would save time for such questions
Fantastic mind blowing sir ❤️👍❤️...
We could just look at the factors of constant -36 (rational root theorem)
Sir please number theory pe lecture series laayiye.
Please sir.please sir.Indian Statistical institute ke admission test ke liye seekhni hai number theory.please sir,please sir,please sir ..............∞
Sir we can assume a^3 as t and then we can use hit and trial
You can use synthetic div method
a⁹+a⁶/2 is greater than or equal to √a⁹a⁶
18² =a¹⁵
a=¹⁵√324
a = 1.47
.
Simple hai, ek aisa number dhoondho jiska cube aur square add karne se 36 aajaye wo number 3 hai, toh 3^1/3
By binomial theorem we can easily get answers
Me hindi midium se hu or mere aise sabal bahut kiye hai masti me
See this easiest solution
Let a^6 =t,
Equation: t^3/2+t=36
. t(√t+1)=36
t=9
a=(9)^1/6
a=(3)^1/3
Destroyed in seconds😂😂😂😂
@@sagarbambal5954 thnkx
@@tanix18 ap maths konsi book se karte ho?
@@sagarbambal5954 Not any specific book.
Only Coaching material and PYQs
a = sixth root of { [-1± sq.root(145)]/2 }
By observation (3)power 1/3😊(my answer)
I don't know how it came to my mind after considering a³ = x ( substitution)
And then x²(x+1)=36
Then x²(x+1) -36 =0
x³+x²-36=0
x³+ (x-6)(x+6) =0
Adding and subtracting 216 ,
x³ + 216+(x-6)(x+6)=216
How did I thought??
My attempt was to convert given question into two simple brackets in product.
Now applying a³+b³ =a+b(a²+b²-ab)
(x+6)(x²+36-6x) +(x-6)(x+6)=216
(x+6)(x²+36-6x +x-6) =216
Now interesting part begins!!
I randomly putted x=6 in above equation and answer came as 432.
Which is twice of 216..
So I thought to try it with x=3 and actually it satisfied.
This is just by God grace. Random thought it was .
I don't know how it came to my mind.
We have substituted a³=x and therefore a³=3
And then a³-3=0
And after applying
a³-b³= (a-b)(a²+b²-ab)
We get (a-3⅓)(a² +3⅔+3⅓a)= 0
Therefore either a=3⅓ or a² +3⅔+3⅓a =0
Clearly we can see that D of Quadratic formed is less than 0 and a( ax²+bx+c) wala is 1 which is greater than 0 and therefore given Quadratic is always greater than 0.
Therefore we get a=3⅓....
I solved it in mind only 😏 cube root 3
Lagta hai dry spell khatm hogya☺️
Me too
Same. It's toooooooo easy.
Sir plz solve this, (a^10) (a+1) = 36 find a
Sir I solved it by myself...
a=3^1/3
Osom quary sir jee
I don't know how but if you practice maths in a regular basis then answer just comes to mind without even doing any calculations
Smj sir very best
How t - 3 is a factor sir pls explain pls pls
Cube root 3
It will be cube root of 3 (3^1/3)
Uhm,
let me tell u how I solved this problem at first glance:-
a^9+a^6=36
1. Take a^6 common
a^6(a^3+1)=36
2. Factorize 36
a^6(a^3+1)=2^2.3^2
3. Now just do some hit and trial by assuming either 2^2 or 3^2 to be a^6
You'll get a=3^1/3 or a=2^1/3
4.Insert the values of a in the original equation and check whether it equates to 36 or not
5. After checking u will get the answer as a=3^1/3.
pls give some hard questions, this was the easiest
Can we do this by using log?
Sir a equal to quberoot 3
You are amazing
a=3^(1/3)
Can we use log?
Wow sir we can solve Olmpird question🙋
a^6(a^3+1)=36
A³=t
t³+t²=36
T=3 satisfy this via hit and trial via this we can find other values too
Answer is cbrt(3)
the answer is cube root of 3.
Ans is (3)^1/3
Of which class is this question
Solved this question
Cubic root of 3
cube root of 3 is the value of a
Cubth root 3
You are great sir💞💞
I did this with in 3 steps
Jhakkas master ji 😃
Qube root of 3 ans
Cube root of 3
I am zygote and i solved this equation
Cube root 3 it took me only 20 second or less than that
Ok genius bhaiya... Atleast ab ye mat bolna ki mai 6vi me hun aur Kota me jee ki star batch me hun, aur nmtc air 1 hun
@@sourthakjackmcober29 main abhi 8th class mein hun tagore school mein
3^1/3=a
Root 3
(3)^1/3
Very nice que
But simple by hit and trial
Ans :√3
Ye easy tha sir atleast mains se to asaan hee laga
Cube rt. 3 in 30 seconds
Legends say he is still spinning in space🌌
cube root of 3 = a
I also solved it. I'm in 2nd
3 to the power 1/3
Sir mera comment aap ke video me aata h class12 ke video bhi banaya karo
mene to by obs se hi krdia a is 3^1/3
3root3
Probably one of the easiest question from Olympiad
3^1/3
This is a class 10 th level question and this mad teacher calling this a
" Olympiad level Question" 🤦
Sir, You should get the title of GENIUS
The third root of 3
Done at first attempt
3 ROOT 3 IS THE ANSWER SIR
I solved it by factoring 36
cube root 2