Simplify the expression (a squared b plus a b squared) times (a squared minus a b plus b squared) minus a b to the fourth power.

Step-by-step solution

1. 1

   Look for structure before expanding. Expanding directly gives six products and is easy to botch. The second bracket $a^2-ab+b^2$ is the tell-tale second factor of the sum of cubes, so the plan is to make the first bracket look like $a+b$ times something.

2. 2

   Factor $ab$ out of the first bracket. Both terms share $ab$:

   $a^2b+ab^2 = ab(a+b)$

   The expression becomes

   $ab(a+b)(a^2-ab+b^2)-ab^4$

3. 3

   Apply the sum-of-cubes identity. Since

   $(a+b)(a^2-ab+b^2)=a^3+b^3$

   the three-bracket product turns into a single simple product:

   $ab(a^3+b^3)-ab^4$

4. 4

   Distribute $ab$.

   $ab \cdot a^3 + ab \cdot b^3 - ab^4 = a^4b + ab^4 - ab^4$

5. 5

   Cancel the opposite terms. $ab^4$ and $-ab^4$ sum to zero, leaving

   $a^4b$

   The fact that the messy $b^4$ piece cancels is the signal that the sum-of-cubes route was the intended one.

6. 6

   Spot-check with numbers. Take $a=2$, $b=3$: the original is $(12+18)(4-6+9)-2\cdot81=30\cdot7-162=210-162=48$, and $a^4b=16\cdot3=48$. Match.