What are the solutions to the equation (3x-2)^3=(3x)^3 ?

The solutions to the equation (3x-2)^3=(3x)^3 are x= 1/3-i(sqrt(3))/9 ,x= 1/3+i(sqrt(3))/9

Find the zeros of (3x-2)^3=(3x)^3

The zeros of (3x-2)^3=(3x)^3 are x= 1/3-i(sqrt(3))/9 ,x= 1/3+i(sqrt(3))/9

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(3 x - 2)^{3} = (3 x)^{3}

x = \frac{1}{3} - i \frac{3}{9}, x = \frac{1}{3} + i \frac{3}{9}

Solution steps

(3 x - 2)^{3} = (3 x)^{3}

Expand

(3 x - 2)^{3} : 27 x^{3} - 54 x^{2} + 36 x - 8

Expand

(3 x)^{3} : 27 x^{3}

27 x^{3} - 54 x^{2} + 36 x - 8 = 27 x^{3}

Move

27 x^{3}

to the left side

- 54 x^{2} + 36 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 ( - 54 ) ( - 8 )}{2 ( - 54 )}

Simplify

3 6^{2} - 4 (- 54) (- 8) : 123 i

x_{1, 2} = \frac{- 36 \pm 12 3 i}{2 ( - 54 )}

Separate the solutions

x_{1} = \frac{- 36 + 12 3 i}{2 ( - 54 )}, x_{2} = \frac{- 36 - 12 3 i}{2 ( - 54 )}

x = \frac{- 36 + 12 3 i}{2 ( - 54 )} : \frac{1}{3} - i \frac{3}{9}

x = \frac{- 36 - 12 3 i}{2 ( - 54 )} : \frac{1}{3} + i \frac{3}{9}

The solutions to the quadratic equation are:

x = \frac{1}{3} - i \frac{3}{9}, x = \frac{1}{3} + i \frac{3}{9}

11 x^{2} - 4 x = - 2

4 x - 8 x^{2} - 16 = - 16

3 x^{2} + 9 x - 13 = 0

y^{2} = 3 y + 70

x^{2} - 4 x + 5 = 0

What are the solutions to the equation (3x-2)^3=(3x)^3 ?
The solutions to the equation (3x-2)^3=(3x)^3 are x= 1/3-i(sqrt(3))/9 ,x= 1/3+i(sqrt(3))/9
Find the zeros of (3x-2)^3=(3x)^3
The zeros of (3x-2)^3=(3x)^3 are x= 1/3-i(sqrt(3))/9 ,x= 1/3+i(sqrt(3))/9