Solve x^6-64b^12

$${ x }^{ 6 } -64 { b }^{ 12 }$$

Factor

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

Solution Steps

Rewrite x^{6}-64b^{12} as \left(x^{3}\right)^{2}-\left(8b^{6}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(x^{3}-8b^{6}\right)\left(x^{3}+8b^{6}\right)$$

Consider x^{3}-8b^{6}. Rewrite x^{3}-8b^{6} as x^{3}-\left(2b^{2}\right)^{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).

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

Consider x^{3}+8b^{6}. Rewrite x^{3}+8b^{6} as x^{3}+\left(2b^{2}\right)^{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).

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

Rewrite the complete factored expression.

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

Evaluate

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