Express $\frac{1}{{{{(1 - {\text{i}}\sqrt 3 )}^3}}}{\text{ in the form }}\frac{a}{b}{\text{ where }}a,{\text{ }}b \in \mathbb{Z}$ .

METHOD 1

$r = 2,{\text{ }}\theta = - \frac{\pi }{3}$     (A1)(A1)

$\therefore {(1 - {\text{i}}\sqrt 3 )^{ - 3}} = {2^{ - 3}}{\left( {\cos \left( { - \frac{\pi }{3}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{3}} \right)} \right)^{ - 3}}$     M1

$= \frac{1}{8}(\cos \pi  + {\text{i}}\sin \pi )$     (M1)

$= - \frac{1}{8}$     A1

[5 marks]

METHOD 2

$(1 - \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 1 - 2\sqrt 3 {\text{i}} - 3\,\,\,\,\,( = - 2 - 2\sqrt 3 {\text{i}})$     (M1)A1

$( - 2 - 2\sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = - 8$     (M1)(A1)

$\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}$     A1

[5 marks]

METHOD 3

Attempt at Binomial expansion     M1

${(1 - \sqrt 3 {\text{i}})^3} = 1 + 3( - \sqrt 3 {\text{i}}) + 3{( - \sqrt 3 {\text{i}})^2} + {( - \sqrt 3 {\text{i}})^3}$     (A1)

$= 1 - 3\sqrt 3 {\text{i}} - 9 + 3\sqrt 3 {\text{i}}$     (A1)

$= - 8$     A1

$\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}$     M1

[5 marks]

Most candidates made a meaningful attempt at this question using a variety of different, but correct methods. Weaker candidates sometimes made errors with the manipulation of the square roots, but there were many fully correct solutions.