Question : If $(p-q)=6,(r-q)=5$ and $(r-p)=3$, then find the value of $\frac{p^3+q^3+r^3-3 p q r}{p+q+r}$.
Option 1: 35
Option 2: 45
Option 3: 30
Option 4: 40
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Correct Answer: 35
Solution :
Given, $(p - q) = 6,$ $(r - q) = 5$ and $(r - p) = 3$
We know, $p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2+ r^2 - pq - pr - qr)$
$⇒\frac{p^3+q^3+r^3-3 p q r}{p+q+r} =\frac{(p + q + r)(p^2 + q^2+ r^2 - pq - pr - qr)}{p+q+r}$
$⇒\frac{p^3+q^3+r^3-3 p q r}{p+q+r} =p^2+q^2+r^2-pq-pr-qr$
$=\frac{1}{2}(2p^2+2q^2+2r^2-2pq-2pr-2qr)$
$=\frac{1}{2}[(p-q)^2+(q-r)^2+(r-p)^2]$
$=\frac{1}{2}[6^2+5^2+3^2]$
$=\frac{1}{2}[36+25+9]$
$=35$
Hence, the correct answer is 35.
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