Solve 729-y^6

$$729 - y ^ { 6 }$$

Factor

$$\left(y-3\right)\left(y+3\right)\left(-y^{2}+3y-9\right)\left(y^{2}+3y+9\right)$$

Solution Steps

Rewrite 729-y^{6} as 27^{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).

$$\left(27+y^{3}\right)\left(27-y^{3}\right)$$

Reorder the terms.

$$\left(y^{3}+27\right)\left(-y^{3}+27\right)$$

Consider y^{3}+27. Rewrite y^{3}+27 as y^{3}+3^{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).

$$\left(y+3\right)\left(y^{2}-3y+9\right)$$

Consider -y^{3}+27. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient -1. One such root is 3. Factor the polynomial by dividing it by y-3.

$$\left(y-3\right)\left(-y^{2}-3y-9\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: -y^{2}-3y-9,y^{2}-3y+9.

$$\left(-y^{2}-3y-9\right)\left(y-3\right)\left(y+3\right)\left(y^{2}-3y+9\right)$$

Evaluate

$$\left(9-y^{2}\right)\left(\left(y^{2}+9\right)^{2}-9y^{2}\right)$$