Factorise $x^{17}-x$ Lets say I have an expression: $$x^{17}-x$$
I have to factorise it.
What I did:
Removed common $x$ outside the expression:$$x(x^{16}-1)$$
What I observed is $x^{16}-1$ can be written as $(x^4)^2-1^2$. So applying the identity of difference of squares $a^2-b^2=(a+b)(a-b)$, I got the answer as $$x(x^4+1)(x^4-1)$$
The last term is a perfect square too. So I again factorised it: $x(x^4+1)(x^2+1)(x^2-1)$
Then Again: $$x(x^4+1)(x^2+1)(x+1)(x-1)$$
But when I used wolfram - alpha to check, I got
$$x(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)$$
Where did I go wrong?
Also, can we write that

*

*$x^4-1$

*$x^2-1$
Are the roots of $x^{17}-x$?
Ideally, if wolfram - alpha gave the result as $(x^8+1)$, so $(x^8-1)$ is also a root if $x^{17}+x$.


Your mistake is in the third equation. In particular you should have: $$x^{16} - 1=(x^8+1)(x^8 - 1)$$
Can you take it from there?
