Solve x^12-64

$${ x }^{ 12 } -64$$

Factor

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

Solution Steps

Rewrite x^{12}-64 as \left(x^{6}\right)^{2}-8^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x^{6}-8\right)\left(x^{6}+8\right)$$

Consider x^{6}-8. Rewrite x^{6}-8 as \left(x^{2}\right)^{3}-2^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

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

Consider x^{6}+8. Rewrite x^{6}+8 as \left(x^{2}\right)^{3}+2^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-2,x^{2}+2,x^{4}-2x^{2}+4,x^{4}+2x^{2}+4.

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

Evaluate

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