# Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** $\sum\alpha' = \alpha\beta + \beta\gamma + \gamma\alpha = -\frac{2}{5}$ | B1 | Quantity must be either identified as, or used as, sum of new roots. NB for reference: $\sum\alpha=\frac{3}{5}$, $\sum\alpha\beta=-\frac{2}{5}$, $\alpha\beta\gamma=-\frac{9}{5}$ |
| $\alpha'\beta'\gamma' = (\alpha\beta)(\beta\gamma)(\gamma\alpha) = (\alpha\beta\gamma)^2\ldots$ | M1 | For expressing product of new roots in terms of old roots |
| $\ldots = (-9/5)^2 = \frac{81}{25}$ | A1 | For expressing product of new roots in terms of old roots. Condone $(9/5)^2$ if seen. Do not condone $-9/5^2$ or $-(9/5)^2$ unless recovered |
| $\sum\alpha'\beta' = (\alpha\beta)(\beta\gamma) + (\beta\gamma)(\gamma\alpha) + (\gamma\alpha)(\alpha\beta) = \alpha\beta\gamma(\alpha+\beta+\gamma)\ldots$ | M1 | Finding sum of products and rewriting into symmetric form |
| $= (-9/5)(3/5) = -\frac{27}{25}$ | A1 | |
| $a=25 \Rightarrow 25x^3 + 10x^2 - 27x - 81 = 0$ | A1 | Or any non-zero integer multiple. Needs to be an equation. |
| **[6]** | | |
| **Alternative method:** $\alpha\beta\gamma = -9/5$ | B1 | |
| $u = \alpha\beta\gamma/x = -9/(5x)$ | B1 | SOI |
| When $x=\alpha$, $u=\beta\gamma$, and similar for other roots | B1 | |
| $5\left(\dfrac{-9}{5u}\right)^3 - 3\left(\dfrac{-9}{5u}\right)^2 - 2\left(\dfrac{-9}{5u}\right) + 9 = 0$ | M1 | |
| $\dfrac{-729}{25u^3} - \dfrac{243}{25u^2} + \dfrac{18}{5u} + 9 = 0$ | M1 | |
| $25x^3 + 10x^2 - 27x - 81 = 0$ | A1 | |

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