What are the solutions to the equation (3x+1)^3=-27 ?

The solutions to the equation (3x+1)^3=-27 are x=-4/3 ,x= 1/6+i(sqrt(3))/2 ,x= 1/6-i(sqrt(3))/2

Find the zeros of (3x+1)^3=-27

The zeros of (3x+1)^3=-27 are x=-4/3 ,x= 1/6+i(sqrt(3))/2 ,x= 1/6-i(sqrt(3))/2

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(3 x + 1)^{3} = - 27

x = - \frac{4}{3}, x = \frac{1}{6} + i \frac{3}{2}, x = \frac{1}{6} - i \frac{3}{2}

Solution steps

(3 x + 1)^{3} = - 27

Expand

(3 x + 1)^{3} : 27 x^{3} + 27 x^{2} + 9 x + 1

27 x^{3} + 27 x^{2} + 9 x + 1 = - 27

Move

27

to the left side

27 x^{3} + 27 x^{2} + 9 x + 28 = 0

Factor

27 x^{3} + 27 x^{2} + 9 x + 28 : (3 x + 4) (9 x^{2} - 3 x + 7)

(3 x + 4) (9 x^{2} - 3 x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 4 = 0 or 9 x^{2} - 3 x + 7 = 0

Solve

3 x + 4 = 0 : x = - \frac{4}{3}

Solve

9 x^{2} - 3 x + 7 = 0 : x = \frac{1}{6} + i \frac{3}{2}, x = \frac{1}{6} - i \frac{3}{2}

The solutions are

x = - \frac{4}{3}, x = \frac{1}{6} + i \frac{3}{2}, x = \frac{1}{6} - i \frac{3}{2}

(1 - x)^{5} = \frac{7}{10}

x^{3} - 64 = 0

x^{5} + x - 1 = 0

x^{3} - 7 x - 6 = 0

(2 + x^{2})^{5} = 1

What are the solutions to the equation (3x+1)^3=-27 ?
The solutions to the equation (3x+1)^3=-27 are x=-4/3 ,x= 1/6+i(sqrt(3))/2 ,x= 1/6-i(sqrt(3))/2
Find the zeros of (3x+1)^3=-27
The zeros of (3x+1)^3=-27 are x=-4/3 ,x= 1/6+i(sqrt(3))/2 ,x= 1/6-i(sqrt(3))/2