What are the solutions to the equation (x+1)^3=(2x)^3 ?

The solutions to the equation (x+1)^3=(2x)^3 are x=1,x=-2/7+i(sqrt(3))/7 ,x=-2/7-i(sqrt(3))/7

Find the zeros of (x+1)^3=(2x)^3

The zeros of (x+1)^3=(2x)^3 are x=1,x=-2/7+i(sqrt(3))/7 ,x=-2/7-i(sqrt(3))/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 1)^{3} = (2 x)^{3}

x = 1, x = - \frac{2}{7} + i \frac{3}{7}, x = - \frac{2}{7} - i \frac{3}{7}

Solution steps

(x + 1)^{3} = (2 x)^{3}

Expand

(x + 1)^{3} : x^{3} + 3 x^{2} + 3 x + 1

Expand

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

x^{3} + 3 x^{2} + 3 x + 1 = 8 x^{3}

Move

8 x^{3}

to the left side

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

Factor

- 7 x^{3} + 3 x^{2} + 3 x + 1 : - (x - 1) (7 x^{2} + 4 x + 1)

- (x - 1) (7 x^{2} + 4 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 7 x^{2} + 4 x + 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

7 x^{2} + 4 x + 1 = 0 : x = - \frac{2}{7} + i \frac{3}{7}, x = - \frac{2}{7} - i \frac{3}{7}

The solutions are

x = 1, x = - \frac{2}{7} + i \frac{3}{7}, x = - \frac{2}{7} - i \frac{3}{7}

x, x^{3} + y^{3} = 8

2000 x^{3} = 1000

x = 33 x^{2} + 4 x^{3} - 10 x^{2}

0.001 x^{3} = - 8

6 = 20 + 12 t^{2} - t^{3}

What are the solutions to the equation (x+1)^3=(2x)^3 ?
The solutions to the equation (x+1)^3=(2x)^3 are x=1,x=-2/7+i(sqrt(3))/7 ,x=-2/7-i(sqrt(3))/7
Find the zeros of (x+1)^3=(2x)^3
The zeros of (x+1)^3=(2x)^3 are x=1,x=-2/7+i(sqrt(3))/7 ,x=-2/7-i(sqrt(3))/7