Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

If $(x^2+\frac{1}{x^2}) = 23, x > 0 $ What is the value of $(x^3+\frac{1}{x^3})$= ?

Options:

140

110

-110 

-140

Correct Answer:

110

Explanation:

If $(x^2+\frac{1}{x^2}) = 23, x > 0 $

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

If $K^2+\frac{1}{K^2}$ = n 

then,  K+ \(\frac{1}{k}\) = \(\sqrt {n + 2}\)

x+ \(\frac{1}{x}\) = \(\sqrt {23 + 2}\) = 5

$(x^3+\frac{1}{x^3})$= 53 - 3 × 5 = 110