题目
Problem
The cubic equation 2x3+5x2−6=0 has roots α, β, γ.
(a) Find a cubic equation whose roots are α31, β31, γ31.
(3)
(b) Find the value of α61+β61+γ61.
(3)
(c) Find also the value of α91+β91+γ91.
(2)
题目中文翻译
三次方程 2x3+5x2−6=0 的根为 α、β、γ。
(a) 求一个以 α31、β31、γ31 为根的三次方程。
(3)
(b) 求 α61+β61+γ61 的值。
(3)
(c) 另求 α91+β91+γ91 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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设新根为 y=x−3。从原方程隔出 x2,立方后用 x3=1/y 消去原变量。
答题过程
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The original roots are nonzero because the constant term is −6. Put y=1/x3. From
2x3+5x2−6=0,
we have 5x2=6−2/y. Cubing gives
125x6=(6−y2)3.
Since x6=1/y2, multiplying by y3 yields
125y==(6y−2)3216y3−216y2+72y−8.
Thus a required cubic equation is
216y3−216y2−53y−8=0.
(b)
解法一
思路
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把(a)的新根记为 p,q,r。原来的六次倒数和就是 p2+q2+r2,由根与系数关系计算。
答题过程
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Let p=α−3, q=β−3 and r=γ−3. From part (a),
p+q+r=1,pq+pr+qr=−21653.
Therefore
===α61+β61+γ61p2+q2+r2(p+q+r)2−2(pq+pr+qr)1+216106=108161.
(c)
解法一
思路
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每个新根都满足(a)的三次方程;把三条等式相加,再代入(b)的平方和。
答题过程
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Each of p,q,r satisfies 216y3=216y2+53y+8. Adding these three equations,
216(p3+q3+r3)==216108161+53(1)+24399.
Hence
=α91+β91+γ91216399=72133.