You may directly factor out `(x + 3)` , such that:

`(x + 3)((x + 3)^2 + 2(x + 3) - 8) = 0 => {(x + 3 = 0),((x + 3)^2 + 2(x + 3) - 8 = 0):}`

`{(x = -3),(x^2 + 6x + 9 + 2x + 6 - 8 = 0):} => {(x = -3),(x^2 + 8x + 7 = 0):}`

You may solve quadratic equation by factorization, such that:

`x^2 + (7 + 1)x + 7*1 = 0`

`x^2 + 7x + x + 7 = 0 => (x^2 + 7x) + (x + 7) = 0`

`x(x + 7) + (x + 7) = 0 => (x + 7)(x + 1) = 0 => {(x + 7 = 0),(x + 1 = 0):} => {(x = -7),(x = -1):}`

**Hence, evaluating the solutions to the given cubic equation, using factorization, yields **`x = -1, x = -7, x = -3.`

We'll replace x + 3 by t.

We'll re-write the equation:

t^3 + 2t^2 - 8t = 0

We'll factorize by t:

t(t^2 + 2t - 8) = 0

We'll cancel each factor:

t = 0

We'll cancel the next factor:

t^2 + 2t - 8 = 0

We'll solve the quadratic equation:

t2 = [-2+sqrt(4+32)]/2

t2 = (-2+6)/2

t2 = 2

t3 = (-2-6)/2

t3 = -4

Let's not forget that we have to calculate x and we did not find it, yet.

x + 3 = t1

x + 3 = 0

x1 = -3

x + 3 = 2

x = 2-3

x2 = -1

x + 3 = -4

x3 = -4-3

x3 = -7

The solutions of the equation are: {-7 , -3 , -1}.