Answer: (i) The roots are \(3,\,6,\,12\).

(ii) \(k=126\).

Let the three real roots be \(a, ar, ar^{-1}\). Then, by Vieta’s relations for

\(x^{3}-21x^{2}+kx-216=0\),

we have

\(a+ar+ar^{-1}=21\),

\(a\cdot ar+ a\cdot ar^{-1}+ ar\cdot ar^{-1}=k\),

and

\(a\cdot ar\cdot ar^{-1}=216\).

From the product,

\(a^{3}=216\), so \(a=6\).

Then the sum of the roots gives

\(6(1+r+r^{-1})=21\), so \(1+r+r^{-1}=\frac{7}{2}\).

Hence

\(r+r^{-1}=\frac{5}{2}\).

Multiplying by \(r\),

\(r^{2}-\frac{5}{2}r+1=0\),

so

\(2r^{2}-5r+2=0\).

Solving,

\(r=2\) or \(r=\frac12\). These give the same set of roots, namely

\(6,\,12,\,3\).

So the numerical values of the roots are \(3,6,12\).

Now use the sum of pairwise products to find \(k\):

\(k=6\cdot 6r + 6\cdot 6r^{-1} + (6r)(6r^{-1})\)

\(=36(r+r^{-1})+36\)

\(=36\cdot \frac{5}{2}+36=90+36=126\).

Therefore \(k=126\).