# Solve the problem (x+3)^3 + 2(x+3)^2 - 8(x+3) = 0 Do i have to turn into an equation about x?

*print*Print*list*Cite

(x+3)^3 + 2(x+3)^2 - 8(x+3) = 0. To solve the equation.

We need not turn into an equation in x , as all the terms are the powers of x+3 and we can treat x+3 itself as one variable t and solve for t.

So we put x+3 = t.

Then the equation changes to t^3+2t^2-8t = 0

We factorise the left.

t(t^2+2t-8) = 0....(1).

Consider t^2+2t-8 for further factorisation:

t^2+2t-8 = t^2+4t-2t - 8

t^2+4t-2t-8 = t(t+4)-2(t+4) = (t+4)(t-2).

Therefore t^2 +2t-8 = (t+4)(t-2).

Substituting t^2+2t-8 = (t+4)(t-2) in eq (1), we get:

t(t+4)(t-2) = 0

Equate each factor to zero:

t = 0 , t+4 = 0 and t-2 = 0.

t = 0 gives t+3 = 0, x = -3.

t+4 = 0 gives x+3+4 = 0, x = -7.

t-2 = 0 gives x+3-2 = 0 , x = -1.

Therefore x = -7, x = -3 or x = -1.

If we'll consider the brackets (x+3) as an entity, we'll solve much easier the equation. So, we'll solve the equation using substitution technique and factorization.

We'll substitute x + 3 = 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 put each factor as zero. We'll start with the first factor t:

t = 0

We'll put the next factor as zero:

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}.**