What are the solutions to the equation 3(x+1)^3=-648 ?

The solutions to the equation 3(x+1)^3=-648 are x=-7,x=2+3sqrt(3)i,x=2-3sqrt(3)i

Find the zeros of 3(x+1)^3=-648

The zeros of 3(x+1)^3=-648 are x=-7,x=2+3sqrt(3)i,x=2-3sqrt(3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 (x + 1)^{3} = - 648

x = - 7, x = 2 + 33 i, x = 2 - 33 i

Solution steps

3 (x + 1)^{3} = - 648

Expand

3 (x + 1)^{3} : 3 x^{3} + 9 x^{2} + 9 x + 3

3 x^{3} + 9 x^{2} + 9 x + 3 = - 648

Move

648

to the left side

3 x^{3} + 9 x^{2} + 9 x + 651 = 0

Factor

3 x^{3} + 9 x^{2} + 9 x + 651 : 3 (x + 7) (x^{2} - 4 x + 31)

3 (x + 7) (x^{2} - 4 x + 31) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 7 = 0 or x^{2} - 4 x + 31 = 0

Solve

x + 7 = 0 : x = - 7

Solve

x^{2} - 4 x + 31 = 0 : x = 2 + 33 i, x = 2 - 33 i

The solutions are

x = - 7, x = 2 + 33 i, x = 2 - 33 i

y^{4} + y = 0

x^{3} - 9 x = 0

(3 + 3 w + 5 w^{2})^{6} = 64

2 x = x^{4}

z^{3} + n = 0

What are the solutions to the equation 3(x+1)^3=-648 ?
The solutions to the equation 3(x+1)^3=-648 are x=-7,x=2+3sqrt(3)i,x=2-3sqrt(3)i
Find the zeros of 3(x+1)^3=-648
The zeros of 3(x+1)^3=-648 are x=-7,x=2+3sqrt(3)i,x=2-3sqrt(3)i