Simplify the expression (a^2 b + a b^2)(a^2 - ab + b^2) - a b^4.

Step-by-step solution

1. 1

   Look for a factoring pattern instead of expanding blindly. Multiplying out directly produces six terms before any cancellation. The second bracket $a^2-ab+b^2$ is the unmistakable signature of the sum-of-cubes identity $(a+b)(a^2-ab+b^2)=a^3+b^3$, so the goal is to make the first bracket look like $a+b$ times something.

2. 2

   Factor $ab$ out of the first bracket.

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

   so the expression becomes

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

3. 3

   Apply the sum of cubes. Since $(a+b)(a^2-ab+b^2)=a^3+b^3$,

   $ab\left(a^3+b^3\right)-ab^4$

   What would have been a six-term expansion is now a two-term product.

4. 4

   Distribute and cancel.

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

   The subtracted term was engineered to cancel exactly one of the two products — that is why the problem collapses to a single monomial.

5. 5

   Check numerically at two points. At $a=2,b=3$: first bracket $=12+18=30$, second $=4-6+9=7$, product $=210$, minus $ab^4=2\cdot81=162$, giving $48$; and $a^4b=16\cdot3=48$ ✓. At $a=-1,b=4$: $(4-16)(1+4+16)-(-1)(256)=(-12)(21)+256=-252+256=4$, and $a^4b=1\cdot4=4$ ✓.