Answer: \(p^{3/2}q^{10/3}r^{-7/3}\); \(x=\frac{1}{27}\) or \(x=\frac18\).

(a) Rewrite each part using powers:

\(\sqrt p=p^{1/2},\qquad (qr^2)^{1/3}=q^{1/3}r^{2/3}.\)

The denominator is

\((q^3p)^{-1}r^3=q^{-3}p^{-1}r^3.\)

Therefore

\(\frac{\sqrt{p}\,(qr^2)^{1/3}}{(q^3p)^{-1}r^3} =\frac{p^{1/2}q^{1/3}r^{2/3}}{p^{-1}q^{-3}r^3}.\)

Subtracting the powers in the denominator gives

\(p^{1/2-(-1)}q^{1/3-(-3)}r^{2/3-3} =p^{3/2}q^{10/3}r^{-7/3}.\)

(b) Let \(u=x^{1/3}\). Then \(x^{2/3}=u^2\), so

\(6u^2-5u+1=0.\)

Factorising,

\((3u-1)(2u-1)=0.\)

Hence

\(u=\frac13\quad\text{or}\quad u=\frac12.\)

Since \(u=x^{1/3}\),

\(x=\left(\frac13\right)^3=\frac1{27} \quad\text{or}\quad x=\left(\frac12\right)^3=\frac18.\)