Solve 64x^6-1

$$64 x ^ { 6 } - 1$$

Factor

$$\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)\left(4x^{2}+2x+1\right)$$

Solution Steps

Rewrite 64x^{6}-1 as \left(8x^{3}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(8x^{3}-1\right)\left(8x^{3}+1\right)$$

Consider 8x^{3}-1. Rewrite 8x^{3}-1 as \left(2x\right)^{3}-1^{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).

$$\left(2x-1\right)\left(4x^{2}+2x+1\right)$$

Consider 8x^{3}+1. Rewrite 8x^{3}+1 as \left(2x\right)^{3}+1^{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).

$$\left(2x+1\right)\left(4x^{2}-2x+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: 4x^{2}-2x+1,4x^{2}+2x+1.

$$\left(2x-1\right)\left(4x^{2}-2x+1\right)\left(2x+1\right)\left(4x^{2}+2x+1\right)$$

Evaluate

$$\left(4x^{2}-1\right)\left(\left(4x^{2}+1\right)^{2}-4x^{2}\right)$$