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

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

A³ = –I.

A^–1 = I – A.

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

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

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

[2 marks]

METHOD 1

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

               $={\left(\gamma \right)}^{12}$     A1

               $={\left({\gamma }^3\right)}^4$       (M1)

               $={\left(-1\right)}^4$

               $=1$     A1

METHOD 2

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

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

[4 marks]

A² = A – I  

⇒ A³ = A² – A      M1A1           

          = A – I – A         A1           

          = –I           AG

Note: Allow other valid methods.

[3 marks]

I = A – A²

A^–1 = A^–1A – A^–1A²        M1A1

⇒ A^–1 = I – A         AG

Note: Allow other valid methods.

[2 marks]