Solve x^6+16x^3+64

$$x ^ { 6 } + 16 x ^ { 3 } + 64$$

Factor

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

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 64. One such factor is x^{3}+8. Factor the polynomial by dividing it by this factor.

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

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

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

Rewrite the complete factored expression. Polynomial x^{2}-2x+4 is not factored since it does not have any rational roots.

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

Evaluate

$$\left(x^{3}+8\right)^{2}$$