Solve a^6-7a^3-8

$$a ^ { 6 } - 7 a ^ { 3 } - 8$$

Factor

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

Solution Steps

Find one factor of the form a^{k}+m, where a^{k} divides the monomial with the highest power a^{6} and m divides the constant factor -8. One such factor is a^{3}-8. Factor the polynomial by dividing it by this factor.

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

Consider a^{3}-8. Rewrite a^{3}-8 as a^{3}-2^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

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

Consider a^{3}+1. Rewrite a^{3}+1 as a^{3}+1^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

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

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

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

Evaluate

$$a^{6}-7a^{3}-8$$