Find y from given information. 
Find $y$ given that $$x^{2}+\frac{1}{x^{2}} = 98;$$ and 
  $$x^{3}+\frac{1}{x^{3}} = y.$$ 

I've got $970$, but that's not the full answer, since I got half  of the points. What else could I get here?

We know,
$(x + \frac 1x)^2 = (x^2 + \frac 1{x^2} + 2.x.\frac 1x)$
$(x + \frac 1x)^2 = (98 + 2)$
$(x + \frac 1x)^2 = 100$
$(x + \frac 1x) = \pm10$
Case 1 - $(x + \frac 1x) > 0$
$(x + \frac 1x) = 10$
Now cube of above equation,
$(x + \frac 1x)^3 = 1000$
$x^3 + \frac 1{x^3} + 3.x.\frac 1x(x + \frac 1x) = 1000$
$x^3 + \frac 1{x^3} + 3(10) = 1000$
$x^3 + \frac 1{x^3} = 1000 - 30$
$x^3 + \frac 1{x^3} = 970$
Also,
$x^3 + \frac 1{x^3} = y$
So y = 970.
Case 2 - $(x + \frac 1x) < 0$
$(x + \frac 1x) = -10$
And we get y = -970.
