Answer: The roots are \(-\frac{7}{2}\), \(-\frac{3}{2}\) and \(\frac{1}{2}\), and the value of \(k\) is \(22\).

Let the roots be \(a-d\), \(a\), \(a+d\).

For \(8x^3+36x^2+kx-21=0\), the sum of the roots is \(-\frac{36}{8}=-\frac{9}{2}\). So

\((a-d)+a+(a+d)=3a=-\frac{9}{2}\), hence \(a=-\frac{3}{2}\).

The product of the roots is \(-\frac{-21}{8}=\frac{21}{8}\), so

\((a-d)\,a\,(a+d)=a(a^2-d^2)=\frac{21}{8}.\)

Substitute \(a=-\frac{3}{2}\):

\(-\frac{3}{2}\left(\frac{9}{4}-d^2\right)=\frac{21}{8}.\)

Multiply by \(-\frac{2}{3}\):

\(\frac{9}{4}-d^2=-\frac{7}{4}\).

So \(d^2=\frac{9}{4}+\frac{7}{4}=4\), giving \(d=2\).

Therefore the roots are

\(a-d=-\frac{3}{2}-2=-\frac{7}{2},\quad a=-\frac{3}{2},\quad a+d=-\frac{3}{2}+2=\frac{1}{2}.\)

Now use the sum of products of the roots taken two at a time. For a monic cubic this equals \(\frac{k}{8}\).

The pairwise products are

\(\left(-\frac{7}{2}\right)\left(-\frac{3}{2}\right)=\frac{21}{4},\quad \left(-\frac{7}{2}\right)\left(\frac{1}{2}\right)=-\frac{7}{4},\quad \left(-\frac{3}{2}\right)\left(\frac{1}{2}\right)=-\frac{3}{4}.\)

So

\(\frac{k}{8}=\frac{21}{4}-\frac{7}{4}-\frac{3}{4}=\frac{11}{4},\)

hence \(k=22\).