Answer: \(y=(2x^3+1)^2\).

Since \(\sqrt{y}\) is plotted against \(x^3\), the straight line has equation

\(\sqrt y=mx^3+c.\)

The line passes through \((2,5)\) and \((10,21)\), so its gradient is

\(m=\dfrac{21-5}{10-2}=\dfrac{16}{8}=2.\)

Use the point \((2,5)\):

\(5=2(2)+c.\)

So

\(c=1.\)

Therefore

\(\sqrt y=2x^3+1.\)

Squaring both sides gives

\(y=(2x^3+1)^2.\)