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

Step-by-step solution

1. 1

   Notice there is no shortcut factor this time. Unlike the variant ending in $2(x+1)^3$, here the last term is $2(x-1)^3$, so $(x+1)$ is not a factor of every term. The reliable route is to expand all four products and collect.

2. 2

   Expand each product.

   $(x+1)^3=x^3+3x^2+3x+1$

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

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

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

3. 3

   Add the cubic coefficients.

   $1-2-1+2=0$

   so the $x^3$ terms vanish and the result is at most quadratic — the same structural surprise as in the companion problem, arrived at differently.

4. 4

   Add the remaining coefficients. For $x^2$: $3-2+1-6=-4$. For $x$: $3+2+1+6=12$. For the constants: $1+2-1-2=0$. Hence

   $-4x^2+12x$

5. 5

   Factor and verify.

   $-4x^2+12x=4x(3-x)=-4x(x-3)$

   At $x=1$: the original is $8-2(4)(0)-2(0)+0=8$, and $4(1)(2)=8\;\checkmark$. At $x=3$: the original is $64-2(16)(2)-4(4)+2(8)=64-64-16+16=0$, and $4(3)(0)=0\;\checkmark$ — the factored form correctly predicts roots at $x=0$ and $x=3$.