Answer: (i) Since \(u=-\alpha+\beta+\gamma\), we have

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

For the cubic \(x^3-x^2-3x-10=0\), the sum of the roots is \(\alpha+\beta+\gamma=1\), so \(u+2\alpha=1\).

Hence \(\alpha=\frac{1-u}{2}\). Substituting this into \(x^3-x^2-3x-10=0\) gives

\(\left(\frac{1-u}{2}\right)^3-\left(\frac{1-u}{2}\right)^2-3\left(\frac{1-u}{2}\right)-10=0\),

which simplifies to \(u^3-u^2-13u+93=0\).

So the cubic with roots \(-\alpha+\beta+\gamma,\ \alpha-\beta+\gamma,\ \alpha+\beta-\gamma\) is \(u^3-u^2-13u+93=0\).

(ii) The product of the roots of \(x^3-x^2-3x-10=0\) is \(\alpha\beta\gamma=10\).

Let \(v\) be one of the new roots, so \(v=\frac{1}{\beta\gamma}\). Then

\(\frac{v}{\alpha}=\frac{1}{\alpha\beta\gamma}=\frac{1}{10}\),

so \(\alpha=10v\).

Substituting \(x=10v\) into the original cubic gives

\((10v)^3-(10v)^2-3(10v)-10=0\),

hence

\(100v^3-10v^2-3v-1=0\).

(i) Let \(u=-\alpha+\beta+\gamma\). Then

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

By Vieta’s formula for \(x^3-x^2-3x-10=0\),

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

Therefore \(u+2\alpha=1\), so

\(\alpha=\frac{1-u}{2}.\)

Now substitute this into the original cubic equation, since \(\alpha\) is a root:

\(\left(\frac{1-u}{2}\right)^3-\left(\frac{1-u}{2}\right)^2-3\left(\frac{1-u}{2}\right)-10=0.\)

Multiply by \(8\):

\((1-u)^3-2(1-u)^2-12(1-u)-80=0.\)

Expand:

\(1-3u+3u^2-u^3-2(1-2u+u^2)-12+12u-80=0.\)

So

\(-u^3+u^2+13u-93=0,\)

hence

\(u^3-u^2-13u+93=0.\)

Thus the required cubic is \(u^3-u^2-13u+93=0\).

(ii) Again by Vieta’s formula,

\(\alpha\beta\gamma=10.\)

Let \(v=\frac{1}{\beta\gamma}\). Then

\(\frac{v}{\alpha}=\frac{1}{\alpha\beta\gamma}=\frac{1}{10},\)

so \(\alpha=10v\).

Substitute \(x=10v\) into the original cubic:

\((10v)^3-(10v)^2-3(10v)-10=0.\)

This gives

\(1000v^3-100v^2-30v-10=0,\)

and dividing by \(10\):

\(100v^3-10v^2-3v-1=0.\)

So the required cubic is \(100v^3-10v^2-3v-1=0\).