Question : If $\left(x^3+\frac{1}{x^3}-k\right)^2+\left(x+\frac{1}{x}-p\right)^2=0$, where k and p are real numbers and $x \neq 0$, then $\frac{k}{p}$ is equal to:
Option 1: ${p}^2+1$
Option 2: ${p}^2+3$
Option 3: ${p}^2-1$
Option 4: $p^2-3$
Correct Answer: $p^2-3$
Solution :
Given:
$(x^3+\frac{1}{x^3}-k)^2+(x+\frac{1}{x}-p)^2=0$
If the sum of the squares of two numbers is zero then both the numbers will also be zero.
⇒ $(x^3+\frac{1}{x^3}-k)^2=0$ and $(x+\frac{1}{x}-p)^2=0$
⇒ $(x^3+\frac{1}{x^3}-k)^2=0$
⇒ $x^3+\frac{1}{x^3} = k$ ----------(i)
Also, $(x+\frac{1}{x}-p)^2=0$
⇒ $x+\frac{1}{x}=p$ ------------(ii)
Now, $(x+\frac{1}{x})^3 = x^3+\frac{1}{x^3} + 3(x+\frac{1}{x})$
⇒ $p^3 = k + 3p$
⇒ $k = p^3 - 3p$
⇒ $k = p(p^2-3)$
⇒ $\frac{k}{p} = p^2-3$
Hence, the correct answer is $p^2-3$.
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