Given that $ x + \frac 1 x = r $ what is the value of: $ x^3 + \frac 1 {x^2}$ in terms of $r$?

Question

Given that $$ x + \cfrac 1 x = r $$

what is the value of: $$ x^3 + \cfrac 1 {x^2}$$ in terms of $r$?

NOTE: it is $\cfrac 1 {x^2}$ and not $ \cfrac 1 {x^3} $

Where I reached so far: $$ \Big(x^3 + \cfrac 1 {x^2}\Big) + \cfrac 1 x \cdot\Big(x^3 + \cfrac 1 {x^2}\Big) = r^3 - r^2 -3r - 2 $$

Any hints??

I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. — Lord_Farin, Jun 19 '13 at 13:23
Moreover, please note that the proper TeX command for ensuring larger fractions is \dfrac, not \cfrac (the canonical use for the latter is in typesetting continued fractions). — Lord_Farin, Jun 19 '13 at 13:24
I was afraid that someone would point out that this was a duplicate and that's why I gave that NOTE in bold!! So I was right! Lol :D — Parth Thakkar, Jun 19 '13 at 13:31

score 2 · Accepted Answer · answered Jun 19 '13 at 13:25

2

If $g$ is a function such that $g(x)=f(x+\frac{1}{x})$ then $g(1/x)=g(x)$.

Your $g(x)=x^3+x^{-2}$ therefore cannot be written as $f(x+1/x)$.

answered Jun 19 '13 at 13:25

Thomas Andrews

But @Michael gives a contradictory answer... – Parth Thakkar Jun 19 '13 at 13:25
Hint: can you solve for $x$ in terms of $r$? – Thomas Andrews Jun 19 '13 at 13:27
But the problem is, I'll get two possible values for x. Now, that'd mean there are 2 possible expressions for $x^3 + x^{-2}$. Doesn't seem to make sense. – Parth Thakkar Jun 19 '13 at 13:28
That's my point - there isn't a single solution. You can get a solution if you allow $\pm$ in your expression. – Thomas Andrews Jun 19 '13 at 13:31
Well then, that's what I'll do - take both solutions! – Parth Thakkar Jun 19 '13 at 13:36

Rasa · Answer 2 · 2013-12-12T18:17:52.993

The answer could be found by direct calculation:

$$ x = \frac{r \pm \sqrt{r^2-4}}{2}$$

$$x^3+\frac{1}{x^2} = -\frac{(- r^2 + r + 1)(r + \sqrt{r^2 - 4} + 2)}{2} \tag{$+$}$$

$$x^3+\frac{1}{x^2} = \frac{4}{(r - \sqrt{r^2 - 4})^2 + (r - \sqrt{r^2 - 4})^{\frac{3}{8}}}\tag{$-$}$$

Edit1 Please note that the domain of $r$ is

$$r\in \{\mathbb{R}-(-2,+2)\}$$

for example, if $r=\cfrac{5}{2}$, then $x=2,\cfrac{1}{2}$ and $$x^3+\frac{1}{x^2}=\frac{33}{4}, \frac{33}{8}$$

score 1 · Answer 3 · answered Jun 19 '13 at 13:24

1

Can you turn the first formula into a quadratic in $x$? Solve that, for $x$ as a function of $r$, and substitute into the second formula.

answered Jun 19 '13 at 13:24

Empy2

Is $x$ a function of $r$? What is $x$ if $r=\frac{5}{2}$? Is $x=2$, or $x=1/2$? – Thomas Andrews Jun 19 '13 at 13:28
No such thing is mentioned. Though I believe that it's like $x$ and $r$ are some constants that adhere to the given equation. – Parth Thakkar Jun 19 '13 at 13:29

3 Answers3