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

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Answer 1 · 2015-05-14T03:14:25

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
$6x^3+1=(6^(1/3)x+1)(6^(2/3)x^2-6^(1/3)x+1)$

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