Find y from given information.

Question

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?

Answer 1

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 $-970$.

Answer 2

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 $y=-970$.

Answer 3

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.

Answer 4

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

$$-10\cdot 98=y-10\Rightarrow x=-970$$