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

The solutions to the equation (x+2)^3=(2x)^3 are x=2,x=-4/7+i(2sqrt(3))/7 ,x=-4/7-i(2sqrt(3))/7

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

The zeros of (x+2)^3=(2x)^3 are x=2,x=-4/7+i(2sqrt(3))/7 ,x=-4/7-i(2sqrt(3))/7

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

x = 2, x = - \frac{4}{7} + i \frac{2 3}{7}, x = - \frac{4}{7} - i \frac{2 3}{7}

Solution steps

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

Expand

(x + 2)^{3} : x^{3} + 6 x^{2} + 12 x + 8

Expand

(2 x)^{3} : 8 x^{3}

x^{3} + 6 x^{2} + 12 x + 8 = 8 x^{3}

Move

8 x^{3}

to the left side

- 7 x^{3} + 6 x^{2} + 12 x + 8 = 0

Factor

- 7 x^{3} + 6 x^{2} + 12 x + 8 : - (x - 2) (7 x^{2} + 8 x + 4)

- (x - 2) (7 x^{2} + 8 x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or 7 x^{2} + 8 x + 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

7 x^{2} + 8 x + 4 = 0 : x = - \frac{4}{7} + i \frac{2 3}{7}, x = - \frac{4}{7} - i \frac{2 3}{7}

The solutions are

x = 2, x = - \frac{4}{7} + i \frac{2 3}{7}, x = - \frac{4}{7} - i \frac{2 3}{7}

- x^{3} + x^{2} = - 2

x^{3} + 2 x^{2} + 4 x - 1 = 0

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

m^{3} - 7 m + 6 = 0

\frac{x ^{4}}{3} = 8

What are the solutions to the equation (x+2)^3=(2x)^3 ?
The solutions to the equation (x+2)^3=(2x)^3 are x=2,x=-4/7+i(2sqrt(3))/7 ,x=-4/7-i(2sqrt(3))/7
Find the zeros of (x+2)^3=(2x)^3
The zeros of (x+2)^3=(2x)^3 are x=2,x=-4/7+i(2sqrt(3))/7 ,x=-4/7-i(2sqrt(3))/7