Simplify E = (x + 2)³ + (x − 2)³ + x³ − 3x(x + 2)(x − 2).

Step-by-step solution

1. 1

   Expand the two cubes with the binomial formula. Using $(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$ with $b = 2$:

   $(x+2)^3 = x^3 + 6x^2 + 12x + 8$

   $(x-2)^3 = x^3 - 6x^2 + 12x - 8$

   The odd-power terms keep their sign pattern, which is what makes the next step collapse.

2. 2

   Add the two cubes — the even-power terms cancel.

   $(x+2)^3 + (x-2)^3 = 2x^3 + 24x$

   The $\pm 6x^2$ terms cancel and so do the $\pm 8$ constants, because they sit at even and odd positions with opposite signs.

3. 3

   Simplify the product term using difference of squares. Deal with the bracketed pair first:

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

   so

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

   Distributing the $-3x$ across both terms (including the $-4$) is where a sign is most often lost.

4. 4

   Collect everything. Assemble the three pieces:

   $E = (2x^3 + 24x) + x^3 + (-3x^3 + 12x)$

   Cubic column: $2x^3 + x^3 - 3x^3 = 0$. Linear column: $24x + 12x = 36x$. There are no $x^2$ or constant terms left.

5. 5

   State and check the result.

   $E = 36x$

   A numeric check at $x = 1$: $27 + (-1) + 1 - 3(3)(-1) = 27 - 1 + 1 + 9 = 36 = 36 \cdot 1$. At $x = 2$: $64 + 0 + 8 - 6(4)(0) = 72 = 36 \cdot 2$. Both confirm the linear result.