Answer
$a^{\frac{3}{4}}$
Work Step by Step
RECALL:
(i) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
(ii) $\dfrac{a^m}{a^n}=a^{m-n}, a\ne0$
(iii) $(am)^m = a^mb^m$
(iv) $(a^m)^n=a^{mn}$
(v) $a^0=1, a\ne0$
Use rule (i) above to obtain:
$=\dfrac{(a^3b)^{\frac{1}{2}}}{(a^3b^2)^{\frac{1}{4}}}$
Use rule (iii) above to obtain:
$=\dfrac{(a^3)^{\frac{1}{2}}b^{\frac{1}{2}}}{(a^3)^{\frac{1}{4}}(b^2)^{\frac{1}{4}}}$
Use rule (iv) above to obtain:
$=\dfrac{a^{3\cdot\frac{1}{2}}b^{\frac{1}{2}}}{a^{3\cdot\frac{1}{4}}b^{2\cdot\frac{1}{4}}}
\\=\dfrac{a^{\frac{3}{2}}b^{\frac{1}{2}}}{b^{\frac{3}{4}}b^{\frac{2}{4}}}
\\=\dfrac{a^{\frac{3}{2}}b^{\frac{1}{2}}}{b^{\frac{3}{4}}b^{\frac{1}{2}}}
$
Use rule (ii) above to obtain:
$=a^{(\frac{3}{2}-\frac{3}{4})}b^{(\frac{1}{2}-\frac{1}{2})}
\\=a^{(\frac{6}{4}-\frac{3}{4})}b^{0}
\\=a^{\frac{3}{4}}b^0$
Use rule (v) above to obtain:
$\\=a^{\frac{3}{4}}\cdot 1
\\=a^{\frac{3}{4}}$
