prove that $\displaystyle
2a\sqrt {\frac{1}
{{6 + 4\sqrt 3 }}} = a\left( {3^{1/4} - 3^{ - 1/4} } \right)
$
Hello, mms!
This requires Olympic-level gymnastics . . .
The left side is: .$\displaystyle \frac{2a}{\sqrt{6 + 4\sqrt{3}}} $ .[1]Prove that: .$\displaystyle 2a\sqrt {\frac{1}
{6 + 4\sqrt 3 }} \:=\: a\left(3^{\frac{1}{4}} - 3^{ -\frac{1}{4}}\right)
$
Note that: .$\displaystyle 6 + 4\sqrt{3} \:=\:\sqrt{3}(2\sqrt{3}+4) \;=\;\sqrt{3}(\sqrt{3}+1)^2$
. . . . . . Hence: .$\displaystyle \sqrt{6+4\sqrt{3}} \;=\;\sqrt[4]{3}(\sqrt{3}+1) $
Then [1] becomes: .$\displaystyle \frac{2a}{\sqrt[4]{3}(\sqrt{3}+1)}$
Multiply by $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}-1}\!:\quad \frac{2a}{\sqrt[4]{3}(\sqrt{3}+1)}\cdot\frac{\sqrt{3}-1}{\sqrt{3}-1} \;=\;\frac{2a(\sqrt{3}-1)}{\sqrt[4]{3}\,(2)} \;=\;\frac{a(\sqrt{3}-1)}{\sqrt[4]{3}} $
Therefore: .$\displaystyle \frac{a(3^{\frac{1}{2}} - 1)}{3^{\frac{1}{4}}} \;=\;a\left(3^{\frac{1}{4}} - 3^{-\frac{1}{4}}\right)$
Hello QM deFuturoNote that:
- $\displaystyle (a+b)^2 = a^2 + 2ab + b^2$
and
- $\displaystyle (\sqrt3)^2 = 3$
Then notice that $\displaystyle 2\sqrt3+4 = 3 + 2\sqrt3 + 1 = (\sqrt3)^2 + 2\sqrt3 + 1$.
Put $\displaystyle a = \sqrt3$ and $\displaystyle b = 1$ in the expansion of $\displaystyle (a+b)^2$, and you're there. I'm afraid this sort of cleverness only comes with years of practice. I (and I suspect Soroban also) have been doing these things since Noah was a boy.
Grandad