Simplify (x+1)^3 - 2(x+1)^2(x-1) - (x+1)(x-1)^2 + 2(x+1)^3(x^2 - 9)/(x + 3).

Step-by-step solution

1. 1

   Kill the fraction first. $x^2-9$ is a difference of squares:

   $x^2-9=(x-3)(x+3)$

   so the last term becomes $\dfrac{2(x+1)^3(x-3)(x+3)}{x+3}=2(x+1)^3(x-3)$, valid for $x\neq -3$. Once the denominator is gone the whole expression is a polynomial.

2. 2

   Factor out the common $(x+1)$. Every one of the four terms carries at least one factor of $x+1$:

   $(x+1)\Bigl[(x+1)^2-2(x+1)(x-1)-(x-1)^2+2(x+1)^2(x-3)\Bigr]$

   Pulling it out now keeps the expansion one degree smaller.

3. 3

   Expand the first three terms inside the bracket. Using $(x+1)^2=x^2+2x+1$, $(x+1)(x-1)=x^2-1$, $(x-1)^2=x^2-2x+1$:

   $\left(x^2+2x+1\right)-2\left(x^2-1\right)-\left(x^2-2x+1\right)=-2x^2+4x+2$

   (The $x^2$ terms give $1-2-1=-2$, the $x$ terms give $2+2=4$, the constants give $1+2-1=2$.)

4. 4

   Expand the fourth term.

   $2(x+1)^2(x-3)=2\left(x^2+2x+1\right)(x-3)=2\left(x^3-x^2-5x-3\right)=2x^3-2x^2-10x-6$

5. 5

   Add the two pieces inside the bracket.

   $\left(-2x^2+4x+2\right)+\left(2x^3-2x^2-10x-6\right)=2x^3-4x^2-6x-4=2\left(x^3-2x^2-3x-2\right)$

6. 6

   Write the final factored form and check it.

   $2(x+1)\left(x^3-2x^2-3x-2\right)$

   The cubic has no rational roots ($\pm1,\pm2$ all fail), so it does not factor further over $\mathbb{Q}$. Numerical check at $x=2.3$: the original expression and this form both equal $-48.2658$. $\checkmark$