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CIE 9709 2025 Nov Paper 15 Q5

A Level / CIE / P1

CIE 9709 2025 Nov Paper 15 Paper · Question 5

题目

Problem

Solve the equation

x3−28+27x3=0.x^3-28+\frac{27}{x^3}=0.
[3]
题目中文翻译

解方程

x3−28+27x3=0.x^3-28+\frac{27}{x^3}=0.

解答

解法一

思路

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原式要求 x≠0x\ne0。乘以 x3x^3 消去分母后,可把 x3x^3 看成一个未知数。

答题过程

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For x≠0x\ne0, multiplying by x3x^3 gives x6−28x3+27=0x^6-28x^3+27=0. Let u=x3u=x^3. Then

u2−28u+27=(u−1)(u−27)=0.u^2-28u+27=(u-1)(u-27)=0.

Thus x3=1x^3=1 or x3=27x^3=27, giving x=1x=1 or x=3x=3. Both are in the original domain.