If $a^2 +\frac{1}{a^2}=98, a > 0, $ then the value of $a^3 + \frac{1}{a^3}$ will be :

970 

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