Factorize ${z^3} + 1$ into a linear and quadratic factor.

Let $\gamma = \frac{{1 + {\text{i}}\sqrt 3 }}{2}$.

(i)     Show that $\gamma$ is one of the cube roots of −1.

(ii)     Show that ${\gamma ^2} = \gamma - 1$.

(iii)     Hence find the value of ${(1 - \gamma )^6}$.

using the factor theorem z +1 is a factor     (M1)

${z^3} + 1 = (z + 1)({z^2} - z + 1)$     A1

[2 marks]

(i)     METHOD 1

${z^3} = - 1 \Rightarrow {z^3} + 1 = (z + 1)({z^2} - z + 1) = 0$     (M1)

solving ${z^2} - z + 1 = 0$     M1

$z = \frac{{1 \pm \sqrt {1 - 4} }}{2} = \frac{{1 \pm {\text{i}}\sqrt 3 }}{2}$     A1

therefore one cube root of −1 is $\gamma$     AG

METHOD 2

${\gamma ^2} = \left( {{{\frac{{1 + i\sqrt 3 }}{2}}^2}} \right) = \frac{{ - 1 + i\sqrt 3 }}{2}$     M1A1

${\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2} \times \frac{{1 + i\sqrt 3 }}{2} = \frac{{ - 1 - 3}}{4}$     A1

= −1     AG

METHOD 3

$\gamma  = \frac{{1 + i\sqrt 3 }}{2} = {e^{i\frac{\pi }{3}}}$     M1A1

${\gamma ^3} = {e^{i\pi }} = - 1$     A1

 

(ii)     METHOD 1

as $\gamma$ is a root of ${z^2} - z + 1 = 0$ then ${\gamma ^2} - \gamma + 1 = 0$     M1R1

$\therefore {\gamma ^2} = \gamma - 1$     AG

Note: Award M1 for the use of ${z^2} - z + 1 = 0$ in any way.

Award R1 for a correct reasoned approach.

METHOD 2

${\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2}$     M1

$\gamma - 1 = \frac{{1 + i\sqrt 3 }}{2} - 1 = \frac{{ - 1 + i\sqrt 3 }}{2}$     A1

 

(iii)     METHOD 1

${(1 - \gamma )^6} = {( - {\gamma ^2})^6}$     (M1)

$= {(\gamma )^{12}}$     A1

$= {({\gamma ^3})^4}$     (M1)

$= {( - 1)^4}$

$= 1$     A1

METHOD 2

${(1 - \gamma )^6}$

$= 1 - 6\gamma + 15{\gamma ^2} - 20{\gamma ^3} + 15{\gamma ^4} - 6{\gamma ^5} + {\gamma ^6}$     M1A1

Note: Award M1 for attempt at binomial expansion.

 

use of any previous result e.g. $= 1 - 6\gamma + 15{\gamma ^2} + 20 - 15\gamma  + 6{\gamma ^2} + 1$     M1

= 1     A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.

 

[9 marks]

In part a) the factorisation was, on the whole, well done.

Part (b) was done well by most although using a substitution method rather than the result above. This used much m retime than was necessary but was successful. A number of candidates did not use the previous results in part (iii) and so seemed to not understand the use of the ‘hence’.