# fun little radical cancellation thingy

Starting with $\sqrt[3]{a} +1 = \sqrt[2]{a}$:

$a + 3\sqrt[3]{a^{2}} + 3\sqrt[3]{a} + 1 = \sqrt[2]{a^{3}}$

Substituting $\sqrt[2]{a}-1$ for $\sqrt[3]{a}$

I have no idea why the rendering is broken for this line, will fix later:

$a+3[\sqrt[2]{a}-1]^{2} + 3[\sqrt[2]{a}-1] + 1 = \sqrt[2]{a^{3}}$

$a + 3[a-2\sqrt[2]{a} +1] + 3\sqrt[2]{a} -3 +1 = \sqrt[2]{a^{3}}$

$a + 3a - 6\sqrt[2]{a} +3 + 3\sqrt[2]{a} -3 + 1 = \sqrt[2]{a^{3}}$

$4a - 3\sqrt[2]{a} + 1 = \sqrt[2]{a^{3}}$

$4a +1 = \sqrt[2]{a^{3}} + 3\sqrt[2]{a}$

What’s amazing here is that if we square both sides, we completely

viz $\sqrt[2]{a^3}\sqrt[2]{a} = \sqrt[2]{a^{4}} = a^{2}$

$16a^{2} +8a + 1 = a^{3} + 6a^{2} +9a$

And you get a single polynomial from this:

$a^{3} -10a^{2} +a -1$

throwing that into wolfram alpha, we find that

$a = \frac{1}{3}\left(10+\sqrt[3]{\frac{1937-33\sqrt[2]{93}}{2}}+\sqrt[3]{\frac{1937+33\sqrt[2]{93}}{2}}\right)$