Solve a^{3}b^3+a^2b^4-ab-b^2

$$a ^ { 3 } b ^ { 3 } + a ^ { 2 } b ^ { 4 } - a b - b ^ { 2 }$$

Factor

$$b\left(a+b\right)\left(ab-1\right)\left(ab+1\right)$$

Solution Steps

Factor out b.

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

Consider a^{3}b^{2}+a^{2}b^{3}-a-b. Do the grouping a^{3}b^{2}+a^{2}b^{3}-a-b=\left(a^{3}b^{2}+a^{2}b^{3}\right)+\left(-a-b\right), and factor out a^{2}b^{2} in the first and -1 in the second group.

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

Factor out common term a+b by using distributive property.

$$\left(a+b\right)\left(a^{2}b^{2}-1\right)$$

Consider a^{2}b^{2}-1. Rewrite a^{2}b^{2}-1 as \left(ab\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(ab-1\right)\left(ab+1\right)$$

Rewrite the complete factored expression.

$$b\left(a+b\right)\left(ab-1\right)\left(ab+1\right)$$

Evaluate

$$b\left(a+b\right)\left(\left(ab\right)^{2}-1\right)$$