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