Express $\frac{\sqrt3}{1 + 2^{1/3}}$ as $\sqrt{2^{2/3} - 1}$

Question

As the title says, I'm trying to show that $\frac{\sqrt3}{1 + 2^{1/3}}$ is equal to $\sqrt{2^{2/3} - 1}$. It's used (without proof) in the solution to a question I'm stuck on, and I've checked they really are the same on Wolfram Alpha, but I can't work out how I'd be able to convert the former into the latter.

Any help you could offer would be really appreciated.

Answer 1

Use that $$3=(2^{2/3}-1)(2^{2/3}+2\times 2^{1/3}+1)$$

Answer 2

Note that\begin{align}\frac3{\left(1+2^{1/3}\right)^2}&=\frac3{1+2\times2^{1/3}+2^{2/3}}\\&=\frac3{1+2^{2/3}+2^{4/3}}\\&=\frac{3\left(1-2^{2/3}\right)}{\left(1+2^{2/3}+2^{4/3}\right)\left(1-2^{2/3}\right)}\\&=\frac{3\left(1-2^{2/3}\right)}{1-4}\\&=2^{2/3}-1.\end{align}

Answer 3

The other answers require you to know our come up with a clever factorisation to get from one expression to the other. However, that's not the only way to show that two expressions represent the same number. Here is a way to see that they are equal without having to be clever about it. Just straight-forward calculation.

Two numbers are equal iff their ratio is $1$. Since they are positive, comparing their squares instead will remove square roots, and make almost everything nicer. We have $$ \frac{2^{1/3}-1}{3/(1+2\cdot 2^{1/3}+1)}\\ =\frac{(2^{1/3}-1)(1+2\cdot 2^{1/3}+1)}3 $$ At this point, just expand the brackets in the numerator, and see that you do indeed get $\frac33=1$ in the end.

Answer 4

Use the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Note: $$\frac{\sqrt3}{1 + 2^{1/3}}=\frac{\sqrt{(2^{1/3})^3+1^3}}{1 + 2^{1/3}}= \frac{\sqrt{(2^{1/3}+1)(2^{2/3}-2^{1/3}+1)}}{2^{1/3}+1}=\\ \sqrt{\frac{2^{2/3}-2^{1/3}+1}{2^{1/3}+1}}=\sqrt{\frac{(2^{1/3}+1)(2^{2/3}-1)}{2^{1/3}+1}}=\sqrt{2^{2/3}-1}.$$