Solve 7s^8-7

$$7 s ^ { 8 } - 7$$

Factor

$$7\left(s-1\right)\left(s+1\right)\left(s^{2}+1\right)\left(s^{4}+1\right)$$

Solution Steps

Factor out 7.

$$7\left(s^{8}-1\right)$$

Consider s^{8}-1. Rewrite s^{8}-1 as \left(s^{4}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider s^{4}-1. Rewrite s^{4}-1 as \left(s^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

$$\left(s-1\right)\left(s+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: s^{2}+1,s^{4}+1.

$$7\left(s-1\right)\left(s+1\right)\left(s^{2}+1\right)\left(s^{4}+1\right)$$

Evaluate

$$7\left(s^{8}-1\right)$$