Solve 1-[(x)^3]^4=

$$1 - [ ( x ) ^ { 3 } ] ^ { 4 } =$$

Factor

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

Solution Steps

Rewrite 1-x^{12} as 1^{2}-\left(-x^{6}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

Consider x^{6}+1. Rewrite x^{6}+1 as \left(x^{2}\right)^{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).

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

Consider -x^{6}+1. Rewrite -x^{6}+1 as 1^{2}-\left(-x^{3}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

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).

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

Consider -x^{3}+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. One such root is 1. Factor the polynomial by dividing it by x-1.

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

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

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

Evaluate

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

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 3 and 4 to get 12.

$$1-x^{12}$$