Question : If $K+\frac{1}{K}=-3$, then what is the value of $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$?
Option 1: 27
Option 2: – 29
Option 3: 29
Option 4: – 27
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Correct Answer: – 29
Solution :
Given equation,
$K+\frac{1}{K}=-3$ ..........(1)
Squaring the equation, we get,
$(K+\frac{1}{K})^2=(-3)^2$
$K^2 + \frac{1}{K^2} + 2 = 9$
⇒ $K^2 + \frac{1}{K^2} = 7$ ...........(2)
Cubing equation (1), we get
$(K+\frac{1}{K})^3 = (-3)^3$
⇒ $K^3 + \frac{1}{K^3} + 3\cdot K\cdot\frac{1}{K}(K+\frac{1}{K}) = -27$
⇒ $K^3+\frac{1}{K^3}+3(-3) = -27$ [Putting value from equation (1)]
⇒ $K^3+\frac{1}{K^3} = -36$ ..........(3)
Now consider, $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$
$= \frac{K^6}{K^3}+\frac{1}{K^3} + \frac{K^4}{K^2}+\frac{1}{K^2}$
$= K^3+\frac{1}{K^3} + K^2+\frac{1}{K^2}$
$=- 36 + 7 = -29$
Hence, the correct answer is – 29.
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