Answer
$7x(x+1)(x-1)(x^2+1)$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $7x^5-7x=\\=7x(x^4-1)\\=7x((x^2)^2-1^2)\\=7x(x^2-1)(x^2+1)\\=7x(x^2-1^2)(x^2+1)\\=7x(x+1)(x-1)(x^2+1)$