Question : The value of $ \frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$, where $p \neq q \neq r$, is equal to:
Option 1: $\frac{1}{9}$
Option 2: $\frac{1}{3}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{1}{2}$
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Correct Answer: $\frac{1}{4}$
Solution :
$\frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$
We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Let $a=p-q$, $b=q-r$ and $c=r-p$
We observe that $a+b+c=p-q+q-r+-p=0$
So, $a^3+b^3+c^3-3abc=0$
or, $a^3+b^3+c^3=3abc$
or, $(p-q)^3+(q-r)^3+(r-p)^3=3(p-q)(q-r)(r-p)$
So, $\frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}= \frac{3(p-q)(q-r)(r-p)}{12(p-q)(q-r)(r-p)}$
= $\frac{1}{4}$
Hence, the correct answer is $\frac{1}{4}$.
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