Answer: \(a=\frac72\), \(b=1\), \(c=\frac16\); \(x=1\) or \(x=\left(\frac53\right)^{5/2}\).

(a) Rewrite each part using powers:

\(\sqrt{pr}=p^{1/2}r^{1/2}\), \(\sqrt[3]{r}=r^{1/3}\), and \((pq)^{-3}=p^{-3}q^{-3}\).

Therefore

Dividing by \(p^{-3}q^{-3}r^{1/3}\) gives

\(p^{1/2+3}q^{-2+3}r^{1/2-1/3} =p^{7/2}q^1r^{1/6}.\)

Hence \(a=\frac72\), \(b=1\), and \(c=\frac16\).

(b) Let

\(u=x^{2/5}\).

Then \(x^{4/5}=u^2\), so the equation becomes

\(3u^2-8u+5=0.\)

Factorising,

\((3u-5)(u-1)=0.\)

Thus

\(u=\frac53\quad\text{or}\quad u=1.\)

Since \(u=x^{2/5}\),

Also \(\left(\frac53\right)^{5/2}\approx3.59\).