Ratio, Proportion, Indices, LogarithmMCQMTP March 22Question 909 of 305
All Questions

If P=x1/3+x−1/3\displaystyle P = x^{1/3} + x^{-1/3} then find value of 3P3−9P\displaystyle 3P^3 - 9P

Options

A3\displaystyle 3
B1/2(x+1/x)\displaystyle 1/2(x+1/x)
C3(x+1/x)\displaystyle 3(x+1/x)
D2(x+1/x)\displaystyle 2(x+1/x)
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Correct Answer

✅ Option c — 3(x+1/x)\displaystyle 3(x+1/x)

All Options:

  • A3\displaystyle 3
  • B1/2(x+1/x)\displaystyle 1/2(x+1/x)
  • C3(x+1/x)\displaystyle 3(x+1/x)
  • D2(x+1/x)\displaystyle 2(x+1/x)

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Detailed Solution & Explanation

We are given:
P=x1/3+x−1/3P = x^{1/3} + x^{-1/3}

Let us cube both sides of the equation using the algebraic identity:
(u+v)3=u3+v3+3uv(u+v)(u+v)^3 = u^3 + v^3 + 3uv(u+v)

Here, let u=x1/3\displaystyle u = x^{1/3} and v=x−1/3\displaystyle v = x^{-1/3}. The product uv\displaystyle uv is:
uv=x1/3×x−1/3=x1/3−1/3=x0=1uv = x^{1/3} \times x^{-1/3} = x^{1/3 - 1/3} = x^0 = 1

Now, applying the identity:
P3=(x1/3)3+(x−1/3)3+3(1)(x1/3+x−1/3)P^3 = \left( x^{1/3} \right)^3 + \left( x^{-1/3} \right)^3 + 3(1)\left( x^{1/3} + x^{-1/3} \right)
P3=x+x−1+3PP^3 = x + x^{-1} + 3P
P3=x+1x+3PP^3 = x + \frac{1}{x} + 3P

Rearranging the terms by moving 3P\displaystyle 3P to the left side:
P3−3P=x+1xP^3 - 3P = x + \frac{1}{x}

We need to find the value of 3P3−9P\displaystyle 3P^3 - 9P. Factor out 3\displaystyle 3 from the expression:
3P3−9P=3(P3−3P)3P^3 - 9P = 3\left( P^3 - 3P \right)

Substituting the value of P3−3P\displaystyle P^3 - 3P:
3P3−9P=3(x+1x)3P^3 - 9P = 3\left( x + \frac{1}{x} \right)

Hence, **Option C** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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