Solve k^135+1

$$k ^ { 135 } + 1$$

Factor

$$\left(k+1\right)\left(k^{4}-k^{3}+k^{2}-k+1\right)\left(k^{10}-k^{5}+1\right)\left(k^{30}-k^{15}+1\right)\left(k^{90}-k^{45}+1\right)$$

Solution Steps

Rewrite k^{135}+1 as \left(k^{45}\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).

$$\left(k^{45}+1\right)\left(k^{90}-k^{45}+1\right)$$

Consider k^{45}+1. Rewrite k^{45}+1 as \left(k^{15}\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).

$$\left(k^{15}+1\right)\left(k^{30}-k^{15}+1\right)$$

Consider k^{15}+1. Rewrite k^{15}+1 as \left(k^{5}\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).

$$\left(k^{5}+1\right)\left(k^{10}-k^{5}+1\right)$$

Consider k^{5}+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 k+1.

$$\left(k+1\right)\left(k^{4}-k^{3}+k^{2}-k+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: k^{4}-k^{3}+k^{2}-k+1,k^{10}-k^{5}+1,k^{30}-k^{15}+1,k^{90}-k^{45}+1.

$$\left(k^{4}-k^{3}+k^{2}-k+1\right)\left(k+1\right)\left(k^{10}-k^{5}+1\right)\left(k^{30}-k^{15}+1\right)\left(k^{90}-k^{45}+1\right)$$

Evaluate

$$k^{135}+1$$