Solve x^12-4096

$$x ^ { 12 } - 4096$$

Factor

$$\left(x-2\right)\left(x+2\right)\left(x^{2}+4\right)\left(x^{2}-2x+4\right)\left(x^{2}+2x+4\right)\left(x^{4}-4x^{2}+16\right)$$

Solution Steps

Rewrite x^{12}-4096 as \left(x^{6}\right)^{2}-64^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x^{6}-64\right)\left(x^{6}+64\right)$$

Consider x^{6}-64. Rewrite x^{6}-64 as \left(x^{3}\right)^{2}-8^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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}+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).

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

Consider x^{6}+64. Rewrite x^{6}+64 as \left(x^{2}\right)^{3}+4^{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).

$$\left(x^{2}+4\right)\left(x^{4}-4x^{2}+16\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-2x+4,x^{2}+2x+4,x^{2}+4,x^{4}-4x^{2}+16.

$$\left(x-2\right)\left(x+2\right)\left(x^{2}-2x+4\right)\left(x^{2}+2x+4\right)\left(x^{2}+4\right)\left(x^{4}-4x^{2}+16\right)$$

Evaluate

$$\left(x^{4}-16\right)\left(x^{4}-4x^{2}+16\right)\left(\left(x^{2}+4\right)^{2}-4x^{2}\right)$$