Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) has a root \(\alpha\), where \(2 < \alpha < 3\).
Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) can be rearranged into the form
$$x ^ { 2 } = 6 - \frac { 1 } { x }$$
(1 mark)
Use the recurrence relation \(x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }\), with \(x _ { 1 } = 2.5\), to find the value of \(x _ { 3 }\), giving your answer to four significant figures.
(2 marks)