Higher November 2018 Paper 2 Q20
20 \(n\) is a positive integer.
Prove algebraically that \(\quad 2n^2\left(\dfrac{3}{n} + n\right) + 6n\left(n^2 - 1\right) \quad\) is a cube number. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{6n^2}{n} + 2n^3\) or \(6n + 2n^3\) or \(6n^3 - 6n\) | M1 | expands one bracket correctly allow \(3 \times 2n\) for \(\dfrac{6n^2}{n}\) |
| \(\dfrac{6n^2}{n} + 2n^3 + 6n^3 - 6n\) or \(6n + 2n^3 + 6n^3 - 6n\) | M1dep | fully correct expansion allow \(3 \times 2n\) for \(\dfrac{6n^2}{n}\) |
| \(8n^3\) and \((2n)^3\) | A1 | must have seen M1M1 oe eg \(8n^3\) and \(2n \times 2n \times 2n\) or \(8n^3\) and \(\sqrt[3]{8n^3} = 2n\) condone \(8n^3\) and \(2^3n^3\) |
Additional guidance
Do not allow \(\quad \dfrac{2n^2 \times 3}{n}\) for \(\dfrac{6n^2}{n}\)