Question : If $x^2-9x+1=0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $54$
Option 2: $108$
Option 3: $702$
Option 4: $810$
Correct Answer: $702$
Solution :
Given: $x^2-9x+1=0$
Dividing both sides by $x$, we get,
$⇒x-9+\frac{1}{x}=0$
$⇒x+\frac{1}{x}=9$
Cubing both sides, we get,
$⇒(x+\frac{1}{x})^3=9^3$
$⇒x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=729$
$⇒x^3+\frac{1}{x^3}+3×9=729$
$\therefore x^3+\frac{1}{x^3}=729-27=702$
Hence, the correct answer is $702$.
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