Question

To find the value of $\sqrt[4]{4-64a^4}$.

Answer 1

$\sqrt[4]{4-64a^4}=\sqrt{\pm 8a^2\sqrt{-1}}$
$2a\sqrt{2}\sqrt{\pm \sqrt{-1}}$
It remains to find the value of $\sqrt{\pm \sqrt{-1}}$
Assume $\sqrt{\pm \sqrt{-1}}=x+y\sqrt{-1}$
then $\pm \sqrt{-1}=x^2-y^2+2xy\sqrt{-1}$

$\therefore x^2-y^2=0$ and $2xy=\pm 1$
whence $x=\frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}$
or $x=-\frac{1}{\sqrt{2}},y=-\frac{1}{\sqrt{2}}$
$\therefore \sqrt{+\sqrt{-1}}=\pm \frac{1}{\sqrt{2}}(1+\sqrt{-1})$
Similarly $\sqrt{-\sqrt{-1}}=\pm \frac{1}{\sqrt{2}}(1-\sqrt{-1})$
$\therefore \sqrt{\pm \sqrt{-1}}=\pm \frac{1}{\sqrt{2}}(1\pm \sqrt{-1})$
and finally $\sqrt[4]{4-64a^4}=\pm 2a(1\pm \sqrt{-1})$