Solve 7x-7x^7

$$7 x - 7 x ^ { 7 }$$

Factor

$$7x\left(x-1\right)\left(x+1\right)\left(-x^{2}+x-1\right)\left(x^{2}+x+1\right)$$

Solution Steps

Factor out 7.

$$7\left(x-x^{7}\right)$$

Consider x-x^{7}. Factor out x.

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

Consider 1-x^{6}. Rewrite 1-x^{6} 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.

$$7x\left(x+1\right)\left(x^{2}-x+1\right)\left(x-1\right)\left(-x^{2}-x-1\right)$$

Evaluate

$$7x\left(1-x^{2}\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)$$