(i)[2]
Show by calculation that $\alpha$ is located between $2$ and $3$.
(ii)[4]
The two iterative formulae, $A$ and $B$, obtained from this equation are as follows: $$x_{n+1} = (3x_n + 7)^{\frac{1}{3}} \; (A)$$ $$x_{n+1} = \frac{x_n^3 - 7}{3} \; (B)$$ Each one is to be used with initial value $x_1 = 2.5$. Show that one of the formulae gives a sequence that does not converge, and use the other formula to find $\alpha$ correct to $2$ decimal places. Give each iterative value to $4$ decimal places.