If \(p+q-r=0\), where \(p\), \(q\) and \(r\) are non-zero, then find the value of \(\displaystyle\frac{p^3+q^3-r^3}{pqr}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-3\)
If three numbers add up to zero, the sum of their cubes equals three times their product: \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\). Here the three numbers are \(p\), \(q\) and \(-r\).
\(p+q+(-r)=0\)
So \(p^3+q^3+(-r)^3=3\cdot p\cdot q\cdot(-r)\), that is \(p^3+q^3-r^3=-3pqr\)
Value \(\displaystyle=\frac{-3pqr}{pqr}=-3\)