If $a^4 +\frac{1}{a^4} = 194$, then what is the value of $\frac{a^6+1}{a^3}$ ?

52

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

According to the question,

$a^4 +\frac{1}{a^4} = 194$

then a² + \(\frac{1}{a^2}\) = \(\sqrt {194 + 2}\) = 14

and a + \(\frac{1}{a}\) = \(\sqrt {14 + 2}\) = 4

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

$\frac{a^6+1}{a^3}$ =  4³ - 3 × 4 = 52