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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


Team Careers360 9th Jan, 2024
Answer (1)
Team Careers360 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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