Solve 9^6-h^6=

$$9 ^ { 6 } - h ^ { 6 } =$$

Factor

$$-\left(h-9\right)\left(h+9\right)\left(h^{2}-9h+81\right)\left(h^{2}+9h+81\right)$$

Solution Steps

Rewrite 531441-h^{6} as 729^{2}-\left(h^{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).

$$\left(729-h^{3}\right)\left(729+h^{3}\right)$$

Reorder the terms.

$$\left(-h^{3}+729\right)\left(h^{3}+729\right)$$

Consider -h^{3}+729. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 729 and q divides the leading coefficient -1. One such root is 9. Factor the polynomial by dividing it by h-9.

$$\left(h-9\right)\left(-h^{2}-9h-81\right)$$

Consider h^{3}+729. Rewrite h^{3}+729 as h^{3}+9^{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).

$$\left(h+9\right)\left(h^{2}-9h+81\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: -h^{2}-9h-81,h^{2}-9h+81.

$$\left(-h^{2}-9h-81\right)\left(h-9\right)\left(h+9\right)\left(h^{2}-9h+81\right)$$

Evaluate

$$\left(81-h^{2}\right)\left(\left(h^{2}+81\right)^{2}-81h^{2}\right)$$

Solution Steps

Calculate 9 to the power of 6 and get 531441.

$$531441-h^{6}$$