Solve a^6-b^6

$$a ^ { 6 } - b ^ { 6 }$$

Factor

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

Solution Steps

Rewrite a^{6}-b^{6} as \left(a^{3}\right)^{2}-\left(b^{3}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Consider a^{3}-b^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

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

Consider a^{3}+b^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

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

Rewrite the complete factored expression.

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

Evaluate

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