Question : If $x+\frac{1}{x}=3, x \neq 0$, then the value of $x^7+\frac{1}{x^7}$ is:
Option 1: 749
Option 2: 843
Option 3: 746
Option 4: 849
Correct Answer: 843
Solution :
If $x+\frac{1}{x}=3$
Using $a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right)$,
$⇒x^3+\frac{1}{x^3} = 3^3 - 3$
$⇒x^3+\frac{1}{x^3} = 18$
Using $a^4 + \frac{1}{a^4} = \left[\left(a + \frac{1}{a}\right)^2 - 2\right]^2 - 2$,
$⇒x^4 + \frac{1}{x^4} = [3^2 - 2]^2 - 2$
$⇒x^4 + \frac{1}{x^4} = 7^2 - 2$
$⇒x^4 + \frac{1}{x^4}= 47$
Using $a^7 + \frac{1}{a^7} = \left(a^3 + \frac{1}{a^3}\right) \times \left(a^4 + \frac{1}{a^4}\right) - \left(a + \frac{1}{a}\right)$,
$⇒x^7+\frac{1}{x^7}=18 \times 47-3$
$⇒x^7+\frac{1}{x^7}= 846 - 3=843$
Hence, the correct answer is 843.
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