Question : If $p+q+r=pqr=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$, find $\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3$.
Option 1: 1
Option 2: –1
Option 3: 5
Option 4: –5
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Correct Answer: 1
Solution : Given: $p+q+r=pqr=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$ $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$ ⇒ $\frac{qr+pr+pq}{pqr}=1$ ⇒ $pr+qr+pq=pqr$ ⇒ $pr+qr+pq=1$ Now, $(p+q+r)^2=1^2$ ⇒ $p^2+q^2+r^2+2(pq+pr+qr)=1$ ⇒ $p^2+q^2+r^2+2=1$ $\therefore p^2+q^2+r^2=-1$ $p^3+q^3+r^3$ $=(p+q+r)(p^2+q^2+r^2-pq-pr-qr)+3pqr$ $=(p+q+r)[p^2+q^2+r^2-(pq+pr+qr)]+3pqr$ $=1×[-1-1]+3×1$ $=-2+3$ $=1$ Hence, the correct answer is 1.
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