Find the product (x + 10)(x^2 - 10x + 100) and simplify your answer.

Step-by-step solution

1. 1

   Recognise the shape before multiplying. The sum-of-cubes identity is

   $(a+b)\left(a^{2}-ab+b^{2}\right)=a^{3}+b^{3}.$

   Match it against the problem: $a=x$ and $b=10$ give $a^2=x^2$, $-ab=-10x$ and $b^2=100$ — every term lines up exactly.

2. 2

   Apply the identity.

   $(x+10)\left(x^{2}-10x+100\right)=x^{3}+10^{3}=x^{3}+1000.$

3. 3

   Verify by expanding the long way. Distribute $x$ and then $10$ across the trinomial:

   $x\left(x^{2}-10x+100\right)=x^{3}-10x^{2}+100x,$
   $10\left(x^{2}-10x+100\right)=10x^{2}-100x+1000.$

4. 4

   Add the two rows and watch the middle terms vanish.

   $x^{3}+\left(-10x^{2}+10x^{2}\right)+\left(100x-100x\right)+1000=x^{3}+1000.$

   The cancellation is not luck — the $-ab$ term in the trinomial is placed precisely so both middle columns cancel.

5. 5

   Spot-check numerically. At $x=1$: $(11)(1-10+100)=11\cdot 91=1001=1^3+1000$. At $x=-10$: the first factor is $0$, and $(-10)^3+1000=0$ as well, confirming that $x=-10$ is the real root of $x^3+1000$.