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?
 A: You could also get $-970$. Your first equation gives four possible values for $x$, yielding only two distinct possible values for $y$, one of which is $-970$.
A: First of all
$$ x^2+\frac{1}{x^2}=\Big (x+\frac{1}{x}\Big)^2-2=98 \iff x+\frac{1}{x}=\pm10$$
Now:  $$y=x^3+\frac{1}{x^3}=\Big (x+\frac{1}{x}\Big)^3-3\Big (x+\frac{1}{x}\Big)=\pm1000-(\pm30)=\pm970$$
Thus: if $\displaystyle x+\frac{1}{x}=10 \Rightarrow y=970,\space$ otherwise if $\displaystyle x+\frac{1}{x}=-10$ then $y=-970$.
A: 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.
A: Although
 I´m little bit late I`m posting an answer. We have
$$ x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=98 $$ $$\left(x+\frac{1}{x}\right)^2=100 $$  $$x+\frac{1}{x}=\pm10$$
Now you can calculate 
$$\left(\frac1x+x \right)\cdot \left(\frac1{x^2}+x^2 \right)=x^3+ \frac1{x^3}+\frac1x+x$$
In case of $x+\frac{1}{x}=10$ we get 
$$10\cdot 98=y+10\Rightarrow x=970$$
In case of $x+\frac{1}{x}=-10$ we get 
$$-10\cdot 98=y-10\Rightarrow x=-970$$
