Question : If $\frac{1}{p}+\frac{1}{q}=\frac{1}{p+q}$, the value of $\left (p^{3}-q^{3}\right )$ is:
Option 1: $p - q$
Option 2: $pq$
Option 3: $1$
Option 4: $0$
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Correct Answer: $0$
Solution : Given: $\frac{1}{p}+\frac{1}{q}=\frac{1}{p+q}$ ⇒ $\frac{q+p}{pq}=\frac{1}{p+q}$ ⇒ $(p+q)^2=pq$ ⇒ $p^2+q^2+2pq=pq$ ⇒ $p^2+q^2+pq=0$ ----------------------------(i) Now, $\left (p^{3}-q^{3}\right )=(p-q)(p^2+q^2+pq)$ Putting the value of $p^2+q^2+pq=0$ from the equation (i) So, $\left (p^{3}-q^{3}\right)=(p-q)(0)$ ⇒ $\left (p^{3}-q^{3}\right)=0$ Hence, the correct answer is $0$.
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