# Ask Uncle Colin: a disguised quadratic

*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions – and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

How do I solve $3x^{\frac{2}{3}} + x^{\frac{1}{3}}-2 = 0$?

I tried cubing everything to get $27x^3 + x - 8 =0$, but I can’t factorise that!

Am I weird?-- Can’t Always Resolve Damned Algebraic Notation

First thing, CARDAN: no, you’re not weird. You’ve just made a mistake. You can’t cube things term by term – if you wanted to cube the whole equation, you’d need to do $\left(3x^{\frac{2}{3}} + x^{\frac{1}{3}}-2\right)^3 = 0^3$, which is a) massive overkill and b) not going to lead to a nice simple solution, I’m afraid.

Don’t worry, though: everyone makes that mistake. Just learn from it and don’t do it again!

The right thing to do is to say $y = x^{1/3}$ – which will turn the equation into a quadratic. You get $3y^2 + y - 2=0$.

This factorises as $(3y-2)(y+1)=0$, so $y=\frac{2}{3}$ or $y=-1$.

But wait! We’re not done. We made $y$ up. $y = x^{\frac{1}{3}}$, so $x^{\frac{1}{3}} = \frac{2}{3}$ or $x^{\frac{1}{3}} = -1$.

*Now* you can cube everything! The first one gives $x = \left( \frac 2 3 \right)^3 = \frac{8}{27}$, and the second gives $x = (-1)^3 = -1$.

Remember: taking powers of fractions is simple; to cube $\frac 23$, you cube the top and cube the bottom.

-- Uncle Colin