Help with a technique in factoring a polynomial of four terms and two variables

Question

I could not simplify this expression into factors despite the time I put into it:

$x^3-64y^3-24xy-8$

I really want to learn how to do this well. Thank you very much for insights/techniques which you may provide...

Answer 1

You can write your expression as $$x^3+(-4y)^3+(-2)^3-3\cdot x \cdot (-4y)\cdot (-2)$$ which is in the form $$s^3+t^3+u^3-3stu$$

This has a standard factorisation which you can derive as follows:

Let $s,t,u$ be the roots of the cubic $y^3-ay^2+by-c=0$ so that $a=s+t+u; b= st+tu+us; c=stu$

Substitute successively $y=s,t,u$ and add to obtain:

$$(s^3+t^3+u^3)-a(s^2+t^2+u^2)+b(s+t+u)-3c=0$$

Using the expressions for $a,b,c$ and reorganising a little, this becomes $$(s^3+t^3+u^3)-3stu=(s+t+u)(s^2+t^2+u^2-st-tu-su)$$

You can apply this in your case to extract one factor. You might also want to note that the quadratic factor satisfies $$2(s^2+t^2+u^2-st-tu-su)=(s-t)^2+(t-u)^2+(u-s)^2$$ so is non-negative for real $s,t,u$. Over the complex numbers take $\omega ^3=1$ a complex cube root of unity, then using $\omega+\omega^2=-1$ $$(s+\omega t +\omega^2 u)(s+\omega^2 t+\omega u)=s^2+t^2+u^2-st-tu-us$$

I've left you to apply this to your own case.