If $x^2 - 3\sqrt{2}x + 1= 0$, then what is the value of $x^3 + (\frac{1}{x^3})$ ?
Answer & explanation
Correct answer: option 2
We know that,
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
If $x^2 - 3\sqrt{2}x + 1= 0$,
then what is the value of $x^3 + (\frac{1}{x^3})$= ?
If $x^2 - 3\sqrt{2}x + 1= 0$,
Divide by x on the both sides of the equation,
x + \(\frac{1}{x}\) = $3\sqrt{2}$
Then, $x^3 +\frac{1}{x^3}$ = ($3\sqrt{2}$)3 - 3 × $3\sqrt{2}$ = $54\sqrt{2}$ - $9\sqrt{2}$ = $45\sqrt{2}$