Simplify (a^2*b + a*b^2)(a^2 - a*b + b^2) - a*b^4.

Step-by-step solution

1. 1

   Look for a familiar identity before multiplying anything out. The second bracket $a^{2}-ab+b^{2}$ is the tell-tale companion of the sum-of-cubes identity

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

   so the goal is to make an $(a+b)$ appear in the first bracket.

2. 2

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

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

   The expression becomes

   $ab(a+b)\left(a^{2}-ab+b^{2}\right)-ab^{4}$

3. 3

   Apply the identity. Replacing $(a+b)\left(a^{2}-ab+b^{2}\right)$ by $a^{3}+b^{3}$ turns a degree-5 expansion into one line:

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

4. 4

   Factor out $ab$ from both remaining terms. Since $ab^{4}=ab\cdot b^{3}$,

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

   The $b^{3}$ terms cancel exactly — that cancellation is the whole reason the problem is posed with $-ab^{4}$.

5. 5

   Combine the powers of $a$.

   $ab\cdot a^{3}=a^{4}b$

6. 6

   Check with a numerical substitution. Take $a=2$, $b=3$: the original is $\left(12+18\right)\left(4-6+9\right)-2\cdot 81=30\cdot 7-162=210-162=48$, and $a^{4}b=16\cdot 3=48\ \checkmark$