Solve a^12-x^8

$$a ^ { 12 } - x ^ { 8 }$$

Factor

$$\left(a^{3}-x^{2}\right)\left(x^{2}+a^{3}\right)\left(x^{4}+a^{6}\right)$$

Solution Steps

Rewrite a^{12}-x^{8} as \left(a^{6}\right)^{2}-\left(x^{4}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{6}-x^{4}\right)\left(a^{6}+x^{4}\right)$$

Reorder the terms.

$$\left(-x^{4}+a^{6}\right)\left(x^{4}+a^{6}\right)$$

Consider -x^{4}+a^{6}. Rewrite -x^{4}+a^{6} as \left(a^{3}\right)^{2}-\left(-x^{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).

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

Reorder the terms.

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

Rewrite the complete factored expression.

$$\left(-x^{2}+a^{3}\right)\left(x^{2}+a^{3}\right)\left(x^{4}+a^{6}\right)$$

Evaluate

$$a^{12}-x^{8}$$