Solve a^{6}+b^{6}-a^2b^4-a^4b^2

$$a ^ { 6 } + b ^ { 6 } - a ^ { 2 } b ^ { 4 } - a ^ { 4 } b ^ { 2 }$$

Factor

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

Solution Steps

Do the grouping a^{6}+b^{6}-a^{2}b^{4}-a^{4}b^{2}=\left(a^{6}-a^{4}b^{2}\right)+\left(-a^{2}b^{4}+b^{6}\right), and factor out a^{4} in the first and -b^{4} in the second group.

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

Factor out common term a^{2}-b^{2} by using distributive property.

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

Consider a^{4}-b^{4}. Rewrite a^{4}-b^{4} as \left(a^{2}\right)^{2}-\left(b^{2}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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