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 3: 3
Option 4: $\frac{1}{3}$
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Correct Answer: $\frac{1}{4}$
Solution : Given: $p=9⇒p^2 = 81$ and $q=\sqrt{17}⇒ q^2= 17$ Putting the values in the equation, we get, $(p^2–q^2)^{–\frac{1}{3}}$ $=(81–17)^{–\frac{1}{3}}$ $=(64)^{–\frac{1}{3}}$ $=(4^3)^{–\frac{1}{3}}$ $=(4)^{–{1}}$ $=\frac{1}{4}$ Hence, the correct answer is $\frac{1}{4}$.
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Question : If $p^2+q^2=7pq$, then the value of $\frac{p}{q}+\frac{q}{p}$ is equal to:
Question : If $\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$ then, find the value of $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}$.
Question : If $c+ \frac{1}{c} =\sqrt{3}$, then the value of $c^{3}+ \frac{1}{c^{3}}$ is equal to:
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:
Question : If $\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}=c$, then the value of $\mathrm{c}$ is:
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