3 The equation \(x ^ { 3 } = 3 x + 7\) has one real root, denoted by \(\alpha\).
- Show by calculation that \(\alpha\) lies between 2 and 3 .
Two iterative formulae, \(A\) and \(B\), derived from this equation are as follows:
$$\begin{aligned}
& x _ { n + 1 } = \left( 3 x _ { n } + 7 \right) ^ { \frac { 1 } { 3 } }
& x _ { n + 1 } = \frac { x _ { n } ^ { 3 } - 7 } { 3 }
\end{aligned}$$
Each formula is used with initial value \(x _ { 1 } = 2.5\). - Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.