Given that \(z ^ { 6 } - 4 z ^ { 3 } + 8 = 0\), show that \(z ^ { 3 } = 2 \pm 2 \mathrm { i }\).
Hence solve the equation
$$z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$$
giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(r > 0\) and \(- \pi < \theta \leqslant \pi\).
Show that, for any real values of \(k\) and \(\theta\),
$$\left( z - k \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z - k \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 2 } - 2 k z \cos \theta + k ^ { 2 }$$
Express \(z ^ { 6 } - 4 z ^ { 3 } + 8\) as the product of three quadratic factors with real coefficients.
8 (a) (i) Given that $z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$, show that $z ^ { 3 } = 2 \pm 2 \mathrm { i }$.\\
(ii) Hence solve the equation
$$z ^ { 6 } - 4 z ^ { 3 } + 8 = 0$$
giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$.\\
(b) Show that, for any real values of $k$ and $\theta$,
$$\left( z - k \mathrm { e } ^ { \mathrm { i } \theta } \right) \left( z - k \mathrm { e } ^ { - \mathrm { i } \theta } \right) = z ^ { 2 } - 2 k z \cos \theta + k ^ { 2 }$$
(c) Express $z ^ { 6 } - 4 z ^ { 3 } + 8$ as the product of three quadratic factors with real coefficients.
\hfill \mbox{\textit{AQA FP2 Q8}}