What are the solutions to the equation (v+4)^3=-27 ?

The solutions to the equation (v+4)^3=-27 are v=-7,v=-5/2+i(3sqrt(3))/2 ,v=-5/2-i(3sqrt(3))/2

Find the zeros of (v+4)^3=-27

The zeros of (v+4)^3=-27 are v=-7,v=-5/2+i(3sqrt(3))/2 ,v=-5/2-i(3sqrt(3))/2

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(v + 4)^{3} = - 27

v = - 7, v = - \frac{5}{2} + i \frac{3 3}{2}, v = - \frac{5}{2} - i \frac{3 3}{2}

Solution steps

(v + 4)^{3} = - 27

Expand

(v + 4)^{3} : v^{3} + 12 v^{2} + 48 v + 64

v^{3} + 12 v^{2} + 48 v + 64 = - 27

Move

27

to the left side

v^{3} + 12 v^{2} + 48 v + 91 = 0

Factor

v^{3} + 12 v^{2} + 48 v + 91 : (v + 7) (v^{2} + 5 v + 13)

(v + 7) (v^{2} + 5 v + 13) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v + 7 = 0 or v^{2} + 5 v + 13 = 0

Solve

v + 7 = 0 : v = - 7

Solve

v^{2} + 5 v + 13 = 0 : v = - \frac{5}{2} + i \frac{3 3}{2}, v = - \frac{5}{2} - i \frac{3 3}{2}

The solutions are

v = - 7, v = - \frac{5}{2} + i \frac{3 3}{2}, v = - \frac{5}{2} - i \frac{3 3}{2}

2 t^{5} + 2 t^{4} - 144 t^{3} = 0

x^{4} + x^{2} - 6 x + 9 = 0

12 x^{3} + 9 x^{2} - 108 x - 81 = 0

(3 x + 5)^{4} - (3 x + 5)^{3} = 0

268 = r^{4}

What are the solutions to the equation (v+4)^3=-27 ?
The solutions to the equation (v+4)^3=-27 are v=-7,v=-5/2+i(3sqrt(3))/2 ,v=-5/2-i(3sqrt(3))/2
Find the zeros of (v+4)^3=-27
The zeros of (v+4)^3=-27 are v=-7,v=-5/2+i(3sqrt(3))/2 ,v=-5/2-i(3sqrt(3))/2