Question : If $pq(p+q)=1$, then the value of $\frac{1}{p^{3}q^{3}}-p^{3}-q^{3}$ is equal to:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4
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Correct Answer: 3
Solution : Given: $pq(p+q)=1$ ⇒ $(p+q)=\frac{1}{pq}$ Cubing both sides, ⇒ $p^3+q^3+3pq(p+q)=\frac{1}{p^3q^3}$ Thus, $\frac{1}{p^3q^3}–p^3–q^3= 3pq(p+q)$ Putting the value of $(p+q)=\frac{1}{pq}$ $= \frac{3}{pq}(pq)$ $=3$ $\therefore\frac{1}{p^3q^3}–p^3–q^3=3$ Hence, the correct answer is 3.
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Question : If $p=9, q=\sqrt{17}$, then the value of $(p^2-q^2)^{-\frac{1}{3}}$ is equal to:
Option 1: 4
Option 2: $\frac{1}{4}$
Option 4: $\frac{1}{3}$
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: $0$
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: $\frac{1}{2}$
Question : If $p^2+q^2=7pq$, then the value of $\frac{p}{q}+\frac{q}{p}$ is equal to:
Option 1: 9
Option 2: 5
Option 3: 7
Option 4: 3
Question : If $P=7+4\sqrt3$ and $PQ=1$, then what is the value of $\frac{1}{P^{2}}+\frac{1}{Q^{2}}$?
Option 1: 196
Option 2: 194
Option 3: 206
Option 4: 182
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