Question : If $x+\frac{1}{x}=7$, find the value of $x^3+\frac{1}{x^3}$.
Option 1: 343
Option 2: 322
Option 3: 340
Option 4: 332
Answer (1)
13th Jan, 2024
Correct Answer: 322
Solution :
$x+\frac{1}{x}=7$
Squaring on both sides,
⇒ $(x+\frac{1}{x})^2=7^2$
⇒ $x^2+\frac{1}{x^2}+2=49$
⇒ $x^2+\frac{1}{x^2}=47$
Now, $x^3+\frac{1}{x^3} = (x+\frac{1}{x})(x^2+\frac{1}{x^2}-(x×\frac{1}{x}))$
= 7 × (47 – 1)
= 322
Hence, the correct answer is 322.
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