Question

How do you factor `6x^3+1`?

Answer 1

There is a useful equality:
`a^n-b^n=`
`=(a-b)*(a^(n-1)b^0+a^(n-2)b^1+a^(n-3)b^2+...+a^2b^(n-3)+a^1b^(n-2)+a^0b^(n-1))`

It can easily be proven by induction.
In case of `n=3` it looks simple:
`a^3-b^3=(a-b)(a^2+ab+b^2)`

Using this equality for `n=3`, `a=6^(1/3)x` and `b=-1`, we get
`6x^3+1=(6^(1/3)x+1)(6^(2/3)x^2-6^(1/3)x+1)`