Answer
$7(x^2+1)(x+1)(x-1)$
Work Step by Step
Factor out the GCF of $7$ to obtain:
$=7(x^4-1)$
The binomial above can be written as $(x^2)^2-1^2$, so the expression above is equivalent to:
$=7[(x^2)^2-1^2]$
Factor the difference of two squares using the formula $a^2-b^2=(a+b)(a-b)$ to obtain:
$=7(x^2+1)(x^2-1)$
Factor the difference of two squares using the same formula to obtain:
$=7(x^2+1)(x+1)(x-1)$
