If you take both sides to the 1/6 power, you get x+2=2^2; x=2. Thereafter, you could expand the expression on the left using Pascal’s triangle, and divide using synthetic division to get the other roots.
Use 1^x to denote exp(2πix) = cos 2πx + i sin 2πx. Lemma: the solutions to x⁶ = 1 are x = 1^{n/6} for n = 0, 1, 2, 3, 4, 5. Proof: at most 6 solutions to a 6th degree polynomial equation can exist. Each of the proposed solutions checks out (just multiply their exponents by 6), and there are 6 of them. Therefore they are all of the solutions. QED. (x + 2)⁶ = 2¹² (= 4⁶) if and only if ((x + 2)/4)⁶ = 1 if and only if (x + 2)/4 = 1^{n/6} for n = 0, 1, 2, 3, 4, 5 (by the lemma), if and only if x = -2 + 4·1^{n/6}, for n = 0, 1, 2, 3, 4, 5. Those are your 6 solutions. In more prosaic and familiar terms: x = -2 + 4(cos πn/3 + i sin πn/3), for n = 0, 1, 2, 3, 4, 5. By the way: x⁶ - 1 = (x - 1^0) (x - 1^⅙) (x - 1^⅓) (x - 1^½) (x - 1^⅔) (x - 1^⅚) = C₁(x) C₂(x) C₃(x) C₆(x), where: C₁(x) = (x - 1^0) = x - 1, C₂(x) = (x - 1^½) = x + 1, C₃(x) = (x - 1^⅓) (x - 1^⅔) = x² + x + 1, C₆(x) = (x - 1^⅙) (x - 1^⅚) = x² - x + 1.
this problem can be solved in your head in 2 minutes. Your solution reminded me of a joke about a gynecologist changing piston rings in a car through the exhaust pipe.
We could reject the complex answers. Simply make it (x + 2) = +/- 4. So x = 2, -6.
If you take both sides to the 1/6 power, you get x+2=2^2; x=2. Thereafter, you could expand the expression on the left using Pascal’s triangle, and divide using synthetic division to get the other roots.
That's a great point! 😊
Use 1^x to denote exp(2πix) = cos 2πx + i sin 2πx. Lemma: the solutions to x⁶ = 1 are x = 1^{n/6} for n = 0, 1, 2, 3, 4, 5. Proof: at most 6 solutions to a 6th degree polynomial equation can exist. Each of the proposed solutions checks out (just multiply their exponents by 6), and there are 6 of them. Therefore they are all of the solutions. QED.
(x + 2)⁶ = 2¹² (= 4⁶) if and only if ((x + 2)/4)⁶ = 1 if and only if (x + 2)/4 = 1^{n/6} for n = 0, 1, 2, 3, 4, 5 (by the lemma), if and only if x = -2 + 4·1^{n/6}, for n = 0, 1, 2, 3, 4, 5. Those are your 6 solutions. In more prosaic and familiar terms: x = -2 + 4(cos πn/3 + i sin πn/3), for n = 0, 1, 2, 3, 4, 5.
By the way:
x⁶ - 1 = (x - 1^0) (x - 1^⅙) (x - 1^⅓) (x - 1^½) (x - 1^⅔) (x - 1^⅚) = C₁(x) C₂(x) C₃(x) C₆(x),
where:
C₁(x) = (x - 1^0) = x - 1,
C₂(x) = (x - 1^½) = x + 1,
C₃(x) = (x - 1^⅓) (x - 1^⅔) = x² + x + 1,
C₆(x) = (x - 1^⅙) (x - 1^⅚) = x² - x + 1.
Essay
this problem can be solved in your head in 2 minutes.
Your solution reminded me of a joke about a gynecologist changing piston rings in a car through the exhaust pipe.
(X+2)⁶ = 2¹²
multiply both side by 1/6 exponentially
= x+2 = 2²
x+2 = 4
x = 2
period!
thanks
X=3 by just looking the equation
Are you serious? 5^6 = 2^12?
(x+2)⁶=2¹²
(X+2)⁶=4⁶ taking 6th on both sides
6√(x+2)⁶=6√4⁶
a√c^b=c^b÷a
(X+2)^(6÷6)=4^(6÷6)
X+2=4
X=4-2
X=2
Nice
x_j=4[cos(jπ/3)+i*sin(jπ/3)]-2, j=0,1,...,5 & i=√(-1)
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[(x+2)³]² - [(2⁶)]² = 0
((x+2)³ - 2⁶)((x+2)³ + 2⁶)) = 0
...so it's a matter of solving these 2 qubic equations:
1) (x+2)³ - (2²)³ = 0
2) (x+2)³ + (2²)³ = 0
1) (x+2)³ - 4³ = 0
[(x+2)-4]([x+2]² + 4² + 4[x+2]) = 0
[x-2](x²+4+4x+ 4² + 4[x+2]) = 0
[x-2](x²+4+4x+ 4² + 4x+8) = 0
(x-2)(x² + 8x + 28) = 0
...this is TEDIOUS but hey I surely can solve it - if you pay me 😂
thanks
@rashel1 No, thank YOU!
At first I thought this was just tedious and boring but it is quite interesting indeed
x_j=4[cos(jπ/3)+i*sin(jπ/3)]-2, j=0,1,...,5 & i=√(-1) for free.
×=2
(x+2)^6=2^12=(2^2)^6=4^6
[{(x+2)/4}^2]^3=1
{(x+2)/4}^2=1 or, w or, w^2= 1 or, w^2 or, (w^3.w) = 1 or, ( w^2 ) or, ( w^4 ).........( where w is the complex cubic root of 1, w=(-1+i(sqrt(3)))/2,,,w^3=1
(x+2)/4 = ( + / - ) 1 or, ( + / - ) w or, ( + / - ) w^2
then , the solution , x = - 2 ( + / - ) 4
or , x = - 2 ( + / - ) 4w
or, x = - 2 ( + / - ) 4w^2
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2^12 = 4^6. (2+x)^6 = 4^6
x + 2 = 4. So x = 2. Solved in 3 seconds.
X=2
(X+2)^6=2^6×2=(2^2)^6=4^6
X+2=4
X=4-2=2
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X=0
Are you serious? Are you saying that 2^6 = 2^12?
X=2