Question 637641
Let U^1/3 = a.
therefore u = a^3.
Now , u^4/3- 17u^2/3 = u^1/3 * u - 17u^1/3 * u^1/3= -16

= a*a^3 - 17a*a = -16
= a^4 - 17a^2 = -16
=  a^4- 17a^2 + 16 = 0

Let a^2 = x,
Now, x^2 - 17x +16 = 0 is a quadratic equation.

x^2 - 16x - x + 16 = 0
x (x -16) - 1(x - 16) = 0
(x-16)(x-1) = 0
x = 16 or x= 1

But x = a^2

so, a^2 = 16 or a^2 = 1

a = 4 or 1 = u^1/3

u= a^3 = 4^3 = 64 or 1

therefore u = 64 OR u = 1