Solution

If \(x = ay^{\frac{3}{4}}\) and \(z = b \sqrt{(y/x)}\), prove that \(z^6 = cx\), where \(c\) is independent of \(x\), \(y\) and \(z\), stating the value of \(c\) in terms of \(a\) and \(b\).

The rules of indices include

\[(pq)^r = p^rq^r\] \[(p^q)^r = p^{qr}\] \[p^qp^r = p^{q+r}\]

We will apply these rules repeatedly here. We have that

\[z = b \sqrt{y/x} \implies z = b x^{-1/2}y^{1/2} \implies z^6 = b^6y^3x^{-3}.\]

We also have that

\[x = ay^{3/4} \implies x^4=a^4y^3 \implies y^3 = x^4a^{-4}.\]

So \(z^6 = b^6a^{-4}x^4x^{-3} = b^6a^{-4}x\), and so \(c = \dfrac{b^6}{a^4}.\)