CAT 2025

Slot 3

QA

Question & SolutionQues & Sol

If $\left( x^{2}+\frac{1}{x^{2}} \right)=25$ and $x>0$, then the value of $\left( x^{7}+\frac{1}{x^{7}} \right)$ is

1. Concept Used

• Topic: Algebra — Algebraic Identities and Powers

• Formula:

$$ \left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 $$

$$ \left(a+\frac{1}{a}\right)^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) $$

2. Calculation

Given,

$$ x^2+\frac{1}{x^2}=25 $$

Therefore,

$$ \left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=25+2=27 $$

Hence,

$$ x+\frac{1}{x}=\sqrt{27}=3\sqrt{3} $$

Also,

$$ \left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right) $$

Therefore,

$$ x^3+\frac{1}{x^3}=(3\sqrt{3})^3-9\sqrt{3}=72\sqrt{3} $$

Also,

$$ \left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2 $$

Therefore,

$$ x^4+\frac{1}{x^4}=25^2-2=623 $$

Lastly,

$$ \left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right) =x^7+\frac{1}{x^7}+x+\frac{1}{x} $$

Therefore,

$$ x^7+\frac{1}{x^7} =623\times72\sqrt{3}-3\sqrt{3} $$

$$ =44853\sqrt{3} $$

3. Solution

Answer = Option A ✅

The value of x⁷ + 1/x⁷ is 44853√3.