Solve b^3+a^3-ab^2-a^2b

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

Factor

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

Solution Steps

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

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

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

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

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

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

Reorder the terms.

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

Rewrite the complete factored expression.

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

Evaluate

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