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 ${(3x^{\frac{2}{3}}+x^{\frac{1}{3}}-2)}^3=0^3$, 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 $\frac{2}{3}$, you cube the top and cube the bottom.

-- Uncle Colin