Solve w^6-64

$$w ^ { 6 } - 64$$

Factor

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

Solution Steps

Rewrite w^{6}-64 as \left(w^{3}\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).

$$\left(w^{3}-8\right)\left(w^{3}+8\right)$$

Consider w^{3}-8. Rewrite w^{3}-8 as w^{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).

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

Consider w^{3}+8. Rewrite w^{3}+8 as w^{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).

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

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

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

Evaluate

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