If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.

Question

I have the following question:

If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.

Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx-1)^{1.5}$ for $x^3$ in the second equation. Ideas?

Hint: you might want to consider the value of $x + 1/x$, and use it as a bridge. — Xiangxiang Xu, Mar 05 '23 at 04:21
The trick of writing everything in terms of (the known) $x+\dfrac1x$ recurs. See for example here or here. Here I describe the theory. Many other threads cover it as well, but those were the ones Approach0 found quickly. Most of the hits lead to AOPS instead because the trick is often used in (easyish) contest questions. — Jyrki Lahtonen, Mar 05 '23 at 06:12
@JyrkiLahtonen Alright, so whenever I see something in terms of $x^n+\frac{1}{x^n}$, I should use $x+\frac{1}{x}$. Thanks! — Tyrcnex, Mar 05 '23 at 07:21
That's correct! And in some other cases as well. Above I added one wrong link. This was what I intended to use. Sorry about pointing at my own old post, those are just easier to find. — Jyrki Lahtonen, Mar 05 '23 at 08:24

John Omielan · Accepted Answer · 2023-03-05T05:17:57.933

I don't see any way to finish using what you've done, since you're dealing with fractional powers of a polynomial, which are generally hard to work with. Instead, since $x \neq 0$, we can divide both sides by $x$ below to get

$$x^2 + 1 = kx \;\;\; \to \;\;\; x+\frac{1}{x}=k$$

Cubing both sides then gives

$$\begin{equation}\begin{aligned} x^3 + 3x^2\left(\frac{1}{x}\right) + 3x\left(\frac{1}{x}\right)^2 + \frac{1}{x^3} & = k^3 \\ x^3 + 3\left(x + \frac{1}{x}\right) + \frac{1}{x^3} & = k^3 \\ x^3 + 3k + \frac{1}{x^3} & = k^3 \\ x^3 + \frac{1}{x^3} & = k^3 - 3k \end{aligned}\end{equation}$$

score 3 · Answer 2 · edited Mar 05 '23 at 05:39

3

\begin{align*}x^3+\frac{1}{x^3}&=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)\\& =\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^2-3\right]\end{align*}

edited Mar 05 '23 at 05:39

Gary

31,845

answered Mar 05 '23 at 04:23

LuckyJollyMoments

455

score 1 · Answer 3 · answered Mar 05 '23 at 05:49

Alternative approach:

Assuming that the roots of the quadratic equation are $x_1,x_2$, Vieta's formulas imply that

$x_1 \times x_2 = 1.$
$x_1 + x_2 = k.$

Therefore, regardless of whether you use $x_1$ or $x_2$ as the value $x$, you have that

$$x + \frac{1}{x} = k.$$

Then, you have that

$$k^3 = \left[x + \frac{1}{x}\right]^3 = x^3 + \frac{1}{x^3} + 3(x)\left(\frac{1}{x}\right)\left[x + \frac{1}{x}\right] $$

$$= x^3 + \frac{1}{x^3} + 3k \implies $$

$$k^3 - 3k = x^3 + \frac{1}{x^3}.$$

If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.

3 Answers3