(i)     $p =  - (\alpha  + \beta  + \gamma )$;

(ii)     $q = \alpha \beta  + \beta \gamma  + \gamma \alpha$;

(iii)     $c =  - \alpha \beta \gamma$.

It is now given that $p =  - 6$ and $q = 18$ for parts (b) and (c) below.

(i)     In the case that the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma$ form an arithmetic sequence, show that one of the roots is $2$.

(ii)     Hence determine the value of $c$.

In another case the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma$ form a geometric sequence. Determine the value of $c$.

(i)-(iii) given the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma$, we have

${x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )$     M1

$= \left( {{x^2} - (\alpha  + \beta )x + \alpha \beta } \right)(x - \gamma )$     A1

$= {x^3} - (\alpha  + \beta  + \gamma ){x^2} + (\alpha \beta  + \beta \gamma  + \gamma \alpha )x - \alpha \beta \gamma$     A1

comparing coefficients:

$p =  - (\alpha  + \beta  + \gamma )$     AG

$q = (\alpha \beta  + \beta \gamma  + \gamma \alpha )$     AG

$c =  - \alpha \beta \gamma$     AG

[3 marks]

METHOD 1

(i)     Given $- \alpha  - \beta  - \gamma  =  - 6$

And $\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18$

Let the three roots be $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma$

So $\beta  - \alpha  = \gamma  - \beta$     M1

or $2\beta  = \alpha  + \gamma$

Attempt to solve simultaneous equations:     M1

$\beta  + 2\beta  = 6$     A1

$\beta  = 2$     AG

(ii)     $\alpha  + \gamma  = 4$

$2\alpha  + 2\gamma  + \alpha \gamma  = 18$

$\Rightarrow {\gamma ^2} - 4\gamma  + 10 = 0$

$\Rightarrow \gamma  = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}$     (A1)

Therefore $c =  - \alpha \beta \gamma  =  - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 =  - 20$     A1

METHOD 2

(i)     let the three roots be $\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d$     M1

adding roots     M1

to give $3\alpha  = 6$     A1

$\alpha  = 2$     AG

(ii)     $\alpha$ is a root, so ${2^3} - 6 \times {2^2} + 18 \times 2 + c = 0$     M1

$8 - 24 + 36 + c = 0$

$c =  - 20$     A1

METHOD 3

(i)     let the three roots be $\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d$     M1

adding roots     M1

to give $3\alpha  = 6$     A1

$\alpha  = 2$     AG

(ii)     $q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)$     M1

${d^2} =  - 6 \Rightarrow d = \sqrt 6 {\text{i}}$

$\Rightarrow c =  - 20$     A1

[5 marks]

METHOD 1

Given $- \alpha  - \beta  - \gamma  =  - 6$

And $\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18$

Let the three roots be $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma$.

So $\frac{\beta }{\alpha } = \frac{\gamma }{\beta }$     M1

or ${\beta ^2} = \alpha \gamma$

Attempt to solve simultaneous equations:     M1

$\alpha \beta  + \gamma \beta  + {\beta ^2} = 18$

$\beta (\alpha  + \beta  + \gamma ) = 18$

$6\beta  = 18$

$\beta  = 3$     A1

$\alpha  + \gamma  = 3,{\text{ }}\alpha  = \frac{9}{\gamma }$

$\Rightarrow {\gamma ^2} - 3\gamma  + 9 = 0$

$\Rightarrow \gamma  = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}$     (A1)(A1)

Therefore $c =  - \alpha \beta \gamma  =  - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 =  - 27$     A1

METHOD 2

let the three roots be $a,{\text{ }}ar,{\text{ }}a{r^2}$     M1

attempt at substitution of $a,{\text{ }}ar,{\text{ }}a{r^2}$ and $p$ and $q$ into equations from (a)     M1

$6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)$     A1

$18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)$     A1

therefore $3 = ar$     A1

therefore $c =  - {a^3}{r^3} =  - {3^3} =  - 27$     A1

[6 marks]

Total [14 marks]