If $a^2 +\frac{1}{a^2}=98, a > 0, $ then the value of $a^3 + \frac{1}{a^3}$ will be :
Answer & explanation
Correct answer: option 3
If x2 + \(\frac{1}{x^2}\) = n
Then, x + \(\frac{1}{x}\) = \(\sqrt {n + 2}\)
we also know that,
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
$a^2 +\frac{1}{a^2}=98, a > 0, $
$a +\frac{1}{a}$ = \(\sqrt {98 + 2}\) = 10
then the value of $a^3 + \frac{1}{a^3}$ = 103 - 3 × 10
= $a^3 + \frac{1}{a^3}$ = 1000 - 30 = 970