CAIE FP1 2008 June — Question 5 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.2 This is a standard Further Maths transformed roots question requiring systematic application of Vieta's formulas and Newton's identities. While it involves multiple steps (finding power sums, constructing the new equation, then computing α⁶+β⁶+γ⁶), the techniques are well-practiced in FP1 syllabi with no novel insight required. The 'show that' format provides the target equation, reducing problem-solving demand. Moderately above average due to algebraic manipulation complexity and being Further Maths content.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
5 The equation $$x ^ { 3 } + x - 1 = 0$$ has roots \(\alpha , \beta , \gamma\). Show that the equation with roots \(\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 }\) is $$y ^ { 3 } - 3 y ^ { 2 } + 4 y - 1 = 0$$ Hence find the value of \(\alpha ^ { 6 } + \beta ^ { 6 } + \gamma ^ { 6 }\).
AnswerMarks
Uses substitution \(y = x^3\)M1
Obtains \(y + y^{1/3} - 1 = 0\)A1
\(y = (1 - y)^3\)A1
\(\Rightarrow ... \Rightarrow y^3 - 3y^2 + 4y - 1 = 0\) (AG)A1
\(\sum \alpha^6 = (\sum \alpha^3)^2 - 2\sum \beta^3 y^3\)B1
\(= 9 - 8 = 1\)M1A1
OR put \(y = z^{1/2}\) to obtain \(z^3 - z^2 + 10z - 1 = 0\)M1A1
\(\sum \alpha^6 = \text{-coefficient of } z^2, = 1\)A1 
Uses substitution $y = x^3$ | M1 |
Obtains $y + y^{1/3} - 1 = 0$ | A1 |
$y = (1 - y)^3$ | A1 |
$\Rightarrow ... \Rightarrow y^3 - 3y^2 + 4y - 1 = 0$ (AG) | A1 |
$\sum \alpha^6 = (\sum \alpha^3)^2 - 2\sum \beta^3 y^3$ | B1 |
$= 9 - 8 = 1$ | M1A1 |
**OR** put $y = z^{1/2}$ to obtain $z^3 - z^2 + 10z - 1 = 0$ | M1A1 |
$\sum \alpha^6 = \text{-coefficient of } z^2, = 1$ | A1 |
5 The equation

$$x ^ { 3 } + x - 1 = 0$$

has roots $\alpha , \beta , \gamma$. Show that the equation with roots $\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 }$ is

$$y ^ { 3 } - 3 y ^ { 2 } + 4 y - 1 = 0$$

Hence find the value of $\alpha ^ { 6 } + \beta ^ { 6 } + \gamma ^ { 6 }$.

\hfill \mbox{\textit{CAIE FP1 2008 Q5 [7]}}
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