How to find the maximum value of $x^2-\frac {1}{x^2}$ if the real number $x$ satisfies $x^3-\frac{1}{x^3}=\frac {28}{27}$

Question

Suppose the real number $x$ satisfies $x^3-\dfrac{1}{x^3}=\dfrac {28}{27}.$ Find the maximum value of $x^2-\dfrac {1}{x^2}.$

A) $\frac{1}{6}\sqrt{37}$

B) $\frac{1}{9}\sqrt{37}$

C) $\frac{1}{3}\sqrt{37}$

D) $\frac{1}{5}\sqrt{37}$

E) $\frac{1}{2}\sqrt{37}$

Attempt.

Let $x^3=y$, so $y-\dfrac {1}{y}=\dfrac {28}{27}$. Hence $27y^2-28y-27=0$. The determinant is $D=28^2+4\times27\times 27=3700$ so $y=x^3=\dfrac {14\pm 5\sqrt {37}}{27}$.

But I'm stuck here doing this calculation $x=\sqrt[3]{\dfrac {14\pm 5\sqrt {37}}{27}}.$

How do I progress further?

Answer 1

Now that you found the 2 values (2 due to the $\pm$) of $x$ that satisfies the equation $x^3-\frac{1}{x^3}=\frac{28}{27}$, we can plug each of them into the expression $x^2-\frac{1}{x^2}$ and determine which of the 2 yields a greater output.

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For $x=\sqrt[3]{\frac{14+5\sqrt{37}}{27}}$

$\left(\sqrt[3]{\frac{14+5\sqrt{37}}{27}}\right)^2-\frac{1}{\left(\sqrt[3]{\frac{14+5\sqrt{37}}{27}}\right)^2}$

$\approx0.676$

For $x=\sqrt[3]{\frac{14-5\sqrt{37}}{27}}$

$\left(\sqrt[3]{\frac{14-5\sqrt{37}}{27}}\right)^2-\frac{1}{\left(\sqrt[3]{\frac{14-5\sqrt{37}}{27}}\right)^2}$

$\approx-0.676$

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$0.676>-0.676$

Thus the maximum value of $x^2-\frac{1}{x^2}$ is approximately $0.676$.

Answer 2

Sometimes you can be lucky and spot the answer. This does not work in general and there may be other methods.

$\sqrt[3]{27}=3$ so the difficult part is $\sqrt[3]{14+5\sqrt{37}}$

If you are lucky then there are rational $a,b$ with $(a+b\sqrt{37})^3= 14+5\sqrt{37}$ which would give you

• $a^3+111ab^2=14$
• $3a^2b+37b^3=5$

and you might spot that $1+111=8\times 14$ and $3+37=8 \times 5$ (spotting $111=3 \times 37$ is a distraction)

so $a=b=\sqrt[3]{\frac18}=\frac12$ will do,

and dividing by the $3$ at the start you get $$x= \sqrt[3]{\frac {14+ 5\sqrt {37}}{27}}=\frac16 +\frac16\sqrt {37}$$ which will give $$x^2-\frac1{x^2} = \frac19\sqrt{37}\approx 0.67586$$ and is going to be better than the smaller magnitude $\sqrt[3]{\frac {14- 5\sqrt {37}}{27}}=\frac16 -\frac16\sqrt {37}$ when maximising $x^2-\frac1{x^2}$.

Answer 3

It seems perfectly possible to reach a solution with your attempts. However, I would like to use another way that works for us in this answer.

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Recall that the following identity :

$$ \begin{align}\left(x-\frac 1x\right)^3+&3\left(x-\frac 1x\right)-\left(x^3-\frac {1}{x^3}\right)=0\end{align} $$

Substituting $x-\dfrac 1x=u$, then we have :

$$ \begin{align}&u^3+3u-\frac {28}{27}=0\\ \implies &27u^3+27\cdot 3u-28=0\\ \implies &\left(3u\right)^3+27\cdot 3u-28=0\end{align} $$

Again substituting $3u=z$, then :

$$ \begin{align}&z^3+27z-28=0\\ \implies &z^3-1+27(z-1)=0\\ \implies &(z-1)\underbrace{(z^2+z+28)}_{\Delta_z<0}=0\end{align} $$

Since $x\in\mathbb R$, we obtain $z=1$, which yields $u=\dfrac 13$. Thus, we determine that $\color{#c00}{\left(x-\dfrac 1x\right)^2=\dfrac 19}$. Finally, using the identity

$$\left(x-\frac 1x\right)^2+4=\left(x+\frac 1x\right)^2$$

leads to :

$$ \begin{align}\color{#c00}{\left(x+\frac 1x\right)^2}&=4+\dfrac 19=\color{#c00}{\dfrac {37}{9}}\end{align} $$

which implies

$$ \begin{align}\left(x^2-\dfrac {1}{x^2}\right)^2&=\dfrac {37}{9}\cdot\dfrac {1}{9}=\dfrac {37}{81}\end{align} $$

Therefore, we conclude that :

$$ \begin{align}\max\left(x^2-\dfrac {1}{x^2}\right)&=\sqrt{\dfrac {37}{81}}=\color{#c00}{\dfrac {\sqrt {37}}{9}}\thinspace .\end{align} $$

Answer 4

Start with $$\left(28\over 27\right)^2=\left(x^3-\frac{1}{x^3}\right)^2=x^6-2+\frac{1}{x^6}.\tag 1$$ From $\;\;x^3={\dfrac {14\pm 5\sqrt {37}}{27}}\;$ we get $$x^6=\left({\dfrac {14\pm 5\sqrt {37}}{27}}\right)^2=\dfrac{1121\pm 140\sqrt{37}}{27^2}\tag 2$$ and due to $(1)$ and $(2)$ $$\frac{1}{x^6}=\left(28\over 27\right)^2+2-x^6=\dfrac{1121\mp 140\sqrt{37}}{27^2}\tag 3$$ Denote $\color{blue}{m=\left(x^2-\frac{1}{x^2}\right)}.$ We are interested in the maximal $m.$ Now, $$\underbrace{\left(x^2-\frac{1}{x^2}\right)^3}_{m^3}=x^6-3\underbrace{\left(x^2-\frac{1}{x^2}\right)}_m-\frac{1}{x^6},$$ that rewrites with the use of $(2)$ and $(3)$ as $$m^3+3m=\frac{\pm 280\sqrt{37}}{27^2}$$ or equivalently, with $\color{blue}{t=9m},$ $$t^3+243t=\pm 280\sqrt{37}\tag 4$$ The function $f(t)=t^3+243t$ is strictly increasing, hence injective, and the largest value of $t$ corresponds to the largest value of RHS in $(4),$ that is $280\sqrt{37}.$
The values in A-E contain $\sqrt{37}.$ It is tentative to plug it in $(4)$ to see that $${\bf{t^3+243t}}=t(t^2+243)=\sqrt{37}(37+243)={\bf{280\sqrt{37}}}.$$ The only solution is $t=\sqrt{37},$ thus $m=\frac{\sqrt{37}}{9}.$

Answer 5

The answers before already provide constructive steps towards the solution, I will just provide a short solution that is acquainted to the contest-math nature of the problem:

Our first observation is that if $y = x^3$ and $b = 28/27$, then $y^2-by-1=0$ which has two roots $r_1$ and $r_2$ with $r_1r_2=-1$ or $r_1 = -1/r_2$ (and this relationship holds for the roots of $x^3-1/x^3 = 28/27$ as well). Thus, if we let $f(x)=x^2-1/x^2$, then $f(r_1)=-f(r_2)$, so we can just find $|f(x)| = \sqrt{f(x)^2}$ for any suitable $x$ we find.

Now, we observe that $28/27 = 1+1/3^3$ isn't arbitrary, and if $s = 3(x - 1/x)$, the binomial theorem (when computing $(x-1/x)^3$) tells us $$x^3-\frac{1}{x^3}=1+\frac{1}{3^3}\iff s+\frac{s^3}{3^3}=1+\frac{1}{3^3}$$ which makes us observe that $s = 1$ works as a solution, and moreover since $(x+1/x)^2=(x-1/x)^2+4$ and $f(x) = (x - 1/x)(x + 1/x)$ then $$f(x)^2=(s/3)^3((s/3)^2+4)$$ which can be computed to give $|f(x)|=\sqrt{37}/9$