Solve c^6-729

$$c ^ { 6 } - 729$$

Factor

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

Solution Steps

Rewrite c^{6}-729 as \left(c^{3}\right)^{2}-27^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider c^{3}-27. Rewrite c^{3}-27 as c^{3}-3^{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).

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

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

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

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

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

Evaluate

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