# Ask Uncle Colin: Factorising a cubic

Dear Uncle Colin,

How would I go about finding $a$, $b$ and $c$ in $x^3 + y^3 = (x+y)\br{ax^2 + bxy + cy^2}$?

Constants Using Brackets & Its Coefficients

Hi, CUBIC, and thanks for your message!

There are, of course, multiple approaches; the most obvious way is to tackle it by dividing everything out in a grid. I’m not going to do that.

You could also multiply out the right hand side and match coefficients.

There’s a way of doing it more logically (and pretty much equivalently).

- There is only one $x^3$ on the left, and the only way to get an $x^3$ on the right is via $ax^3$, so $a=1$.
- Similarly, there is only one way to get a $y^3$, and so $c=1$.
- You now have $(x+y)(x^2 + bxy + y^2) \equiv x^3 + y^3$
- The only ways to get $x^2y$ terms are via $x(bxy)$ and $yx^2$, so $b=-1$.

Job done.

Alternatively, since this is true for any values of $x$ and $y$, you can simply pick some values:

- Suppose $x = 0$: $y^3 = y(cy^2)$, so $c=1$.
- Suppose $y = 0$: $x^3 = x(ax^2)$, so $a=1$.
- Suppose $x = y$: $2x^3 = 2x\br{a+b+c}x^2$, so $a+b+c=1$ and $b=-1$.

Hope that helps!

- Uncle Colin