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’.