Answer: (i) \((\alpha+1)(\beta+1)(\gamma+1)=\frac{19}{2}\).

(ii) \((\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)=-2\).

For the cubic \(2x^3-3x^2+4x-10=0\) with roots \(\alpha,\beta,\gamma\), compare with Vieta’s formulae for \(ax^3+bx^2+cx+d=0\):

\(\alpha+\beta+\gamma=-\frac{b}{a}=\frac{3}{2},\quad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=2,\quad \alpha\beta\gamma=-\frac{d}{a}=5.\)

(a)

Expand the product:

\((\alpha+1)(\beta+1)(\gamma+1)=\alpha\beta\gamma+(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha+\beta+\gamma)+1.\)

Substitute the values:

\(=5+2+\frac{3}{2}+1=\frac{19}{2}.\)

(b)

Write each factor as \(\frac{3}{2}-\) the remaining root:

\(\beta+\gamma=\left(\alpha+\beta+\gamma\right)-\alpha=\frac{3}{2}-\alpha,\)

and similarly \(\gamma+\alpha=\frac{3}{2}-\beta\), \(\alpha+\beta=\frac{3}{2}-\gamma\).

So

\((\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)=\left(\frac{3}{2}-\alpha\right)\left(\frac{3}{2}-\beta\right)\left(\frac{3}{2}-\gamma\right).\)

Expanding gives

\(\left(\frac{3}{2}\right)^3-\left(\frac{3}{2}\right)^2(\alpha+\beta+\gamma)+\frac{3}{2}(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma.\)

Now substitute the symmetric sums:

\(=\frac{27}{8}-\frac{9}{4}\cdot\frac{3}{2}+\frac{3}{2}\cdot 2-5.\)

Hence

\(=\frac{27}{8}-\frac{27}{8}+3-5=-2.\)