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CIE 9231 2026 June Paper 13 Q1

A Level / CIE / FP1

CIE 9231 2026 June Paper 13 Paper · Question 1

题目

Problem

The cubic equation x3+bx2+cx+d=0x^3+bx^2+cx+d=0, where bb, cc and dd are constants, has roots α\alpha, β\beta and γ\gamma. It is given that

α+β+γ= 5,α2+β2+γ2= −19,α−1+β−1+γ−1= 11.\begin{align*} \alpha+\beta+\gamma=&\,5,\\ \alpha^2+\beta^2+\gamma^2=&\,-19,\\ \alpha^{-1}+\beta^{-1}+\gamma^{-1}=&\,11. \end{align*}

(a) Find the values of bb, cc and dd.

(5)

(b) Find the value of α3+β3+γ3\alpha^3+\beta^3+\gamma^3.

(2)

题目中文翻译

三次方程 x3+bx2+cx+d=0x^3+bx^2+cx+d=0 的根为 α\alpha、β\beta 和 γ\gamma，其中 bb、cc 和 dd 为常数。已知

α+β+γ= 5,α2+β2+γ2= −19,α−1+β−1+γ−1= 11.\begin{align*} \alpha+\beta+\gamma=&\,5,\\ \alpha^2+\beta^2+\gamma^2=&\,-19,\\ \alpha^{-1}+\beta^{-1}+\gamma^{-1}=&\,11. \end{align*}

(a) 求 bb、cc 和 dd 的值。

(5)

(b) 求 α3+β3+γ3\alpha^3+\beta^3+\gamma^3 的值。

(2)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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依次用根和、平方和及倒数和求三个系数。倒数和的分母是三个根的乘积，韦达公式带负号。

答题过程

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By Vieta’s formulae,

α+β+γ=−b=5,b=−5.\alpha+\beta+\gamma=-b=5, \qquad\boxed{b=-5}.

Also αβ+βγ+γα=c\alpha\beta+\beta\gamma+\gamma\alpha=c. Using the sum of squares,

−19= (α+β+γ)2−2c= 25−2c,\begin{align*} -19=&\,(\alpha+\beta+\gamma)^2-2c\\[4mm] =&\,25-2c, \end{align*}

so c=22\boxed{c=22}.

Since αβγ=−d\alpha\beta\gamma=-d, the reciprocal sum gives

11=αβ+βγ+γααβγ=22−d.11=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma} =\frac{22}{-d}.

Thus d=−2\boxed{d=-2}.

解法二

思路

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另一种求常数项的方法是将各根的方程除以该根，再把三式相加；根非零由已给倒数和保证。

答题过程

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The sum of roots gives b=−5b=-5. The sum of squares gives

−19=52−2c,c=22.-19=5^2-2c, \qquad c=22.

For each root z≠0z\ne0, divide z3−5z2+22z+d=0z^3-5z^2+22z+d=0 by zz:

z2−5z+22+dz=0.z^2-5z+22+\frac dz=0.

Summing over the three roots,

−19−5(5)+22(3)+11d=0.-19-5(5)+22(3)+11d=0.

Hence 22+11d=022+11d=0, giving

b=−5,c=22,d=−2.\boxed{b=-5},\qquad\boxed{c=22},\qquad\boxed{d=-2}.

(b)

解法一

思路

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各根都满足原方程，将三式直接相加，常数项要出现三次。

答题过程

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For each root zz,

z3=5z2−22z+2.z^3=5z^2-22z+2.

Summing over α,β,γ\alpha,\beta,\gamma gives

α3+β3+γ3= 5(−19)−22(5)+3(2)= −199.\begin{align*} \alpha^3+\beta^3+\gamma^3 =&\,5(-19)-22(5)+3(2)\\[4mm] =&\,\boxed{-199}. \end{align*}