Solve 8x^6+7x^3-1

$$8 x ^ { 6 } + 7 x ^ { 3 } - 1$$

Factor

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

Solution Steps

Find one factor of the form kx^{m}+n, where kx^{m} divides the monomial with the highest power 8x^{6} and n divides the constant factor -1. One such factor is 8x^{3}-1. Factor the polynomial by dividing it by this factor.

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

Consider 8x^{3}-1. Rewrite 8x^{3}-1 as \left(2x\right)^{3}-1^{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).

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

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+1,4x^{2}+2x+1.

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

Evaluate

$$8x^{6}+7x^{3}-1$$