# Simplify $\left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$

Question:

Simplify

$$\left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$$

Attempted solution:

Rewriting it to look a bit better:

$$\left(ab \sqrt[4]{\frac{a^{3}}{\sqrt{b\sqrt{b}}}}\right)^{2}$$

Rewriting the denominator by replacing square roots with powers of fractions:

$$\left(ab \sqrt[4]{\frac{a^{3}}{(b (b^{\frac{1}{2}}))^{\frac{1}{2}}}}\right)^{2}$$

Combining b:s in the denominator:

$$\left(ab \sqrt[4]{\frac{a^{3}}{b^{\frac{3}{4}}}}\right)^{2}$$

Distributing the 4th root:

$$\left(\frac{ab \cdot a^{\frac{12}{16}}}{b^{\frac{3}{16}}}\right)^{2}$$

Combining a:s with a:s and b:s with b:s:

$$\left(b^{\frac{13}{16}} a^{\frac{28}{16}}\right)^{2}$$

Squaring gives the final result:

$$b^{26} a^{56}$$

The answer turns out to be:

$$a^{\frac{7}{2}} b^{\frac{13}{8}}$$

This is quite far away from the result I reached. Where did I go wrong, and are there are key insights that are useful for solving questions with lots and lots of square, cube and higher roots?

Note that $$\left(b^{\frac{13}{16}}a^{\frac{28}{16}}\right)^2=b^{\frac{13}{16}\cdot 2}\times a^{\frac{28}{16}\cdot 2}.$$
$\left(b^{13\over16}a^{28\over16}\right)^2=b^{26\over16}a^{56\over16}=b^{13\over8}a^{7\over2}$
Because $\left(x^y\right)^z=x^{yz}$
we have $a^2b^2\sqrt{\frac{a^3}{b^{3/4}}}=\sqrt{\frac{a^4b^4a^3}{b^{3/4}}}=a^{7/2}(b^{13/4})^{1/2}=a^{7/2}b^{13/8}$