Solve x^9+3x^3-3x^6-1

$$x ^ { 9 } + 3 x ^ { 3 } - 3 x ^ { 6 } - 1$$

Factor

$$\left(x-1\right)^{3}\left(x^{2}+x+1\right)^{3}$$

Solution Steps

Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{9} and m divides the constant factor -1. One such factor is x^{3}-1. Factor the polynomial by dividing it by this factor.

$$\left(x^{3}-1\right)\left(x^{6}-2x^{3}+1\right)$$

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

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

Consider x^{6}-2x^{3}+1. Find one factor of the form x^{n}+p, where x^{n} divides the monomial with the highest power x^{6} and p divides the constant factor 1. One such factor is x^{3}-1. Factor the polynomial by dividing it by this factor.

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

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

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

Rewrite the complete factored expression. Polynomial x^{2}+x+1 is not factored since it does not have any rational roots.

$$\left(x-1\right)^{3}\left(x^{2}+x+1\right)^{3}$$

Evaluate

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