Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

If x + y + z = 0, then what is the value of $\frac{x}{(yz)^2}+\frac{y}{(xz)^2}+\frac{z}{(xy)^2}$ ?

Options:

$\frac{xyz}{3}$

$\frac{3}{xyz}$

xyz

$\frac{1}{xyz}$

Correct Answer:

$\frac{3}{xyz}$

Explanation:

x + y + z = 0

$\frac{x}{(yz)^2}+\frac{y}{(xz)^2}+\frac{z}{(xy)^2}$

If x + y + z = 0 then

a3 + b3 + c3 = 3abc

$\frac{x}{(yz)^2}+\frac{y}{(xz)^2}+\frac{z}{(xy)^2}$ = \(\frac{x^3 + y^3 + z^3}{(xyz)^2}\)

$\frac{x}{(yz)^2}+\frac{y}{(xz)^2}+\frac{z}{(xy)^2}$ = \(\frac{3xyz}{(xyz)^2}\) 

$\frac{x}{(yz)^2}+\frac{y}{(xz)^2}+\frac{z}{(xy)^2}$ = $\frac{3}{xyz}$