Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{9}$
Correct Answer: $\frac{1}{2}$
Solution :
Given expression, $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$
Let $x-y = a,$ $y-z = b,$ and $z-x = c$
⇒ The given expression becomes, $\frac{a^3+b^3+c^3}{6abc}$
Now, $a+b+c= x-y+y-z+z-x = 0$
⇒ When $a+b+c=0,\ a^3 + b^3 + c^3 = 3abc$
⇒ $\frac{a^3+b^3+c^3}{6abc} = \frac{3abc}{6abc}=\frac{1}{2}$
Hence, the correct answer is $\frac{1}{2}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
