Solve 64-y^6

$$64 - y ^ { 6 }$$

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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

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

Consider -y^{3}+8. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient -1. One such root is 2. Factor the polynomial by dividing it by y-2.

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

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

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

Evaluate

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