Solve x^6-7x^3-8

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

Factor

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

Solution Steps

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

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

Consider x^{3}-8. Rewrite x^{3}-8 as x^{3}-2^{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).

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

Consider x^{3}+1. Rewrite x^{3}+1 as x^{3}+1^{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).

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

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

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

Evaluate

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