## Tuesday, August 4, 2015

### IMO Solving Equation Problem: Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real (First Solution)

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where $a$ is real.

If one wants to do things in haste and quickly solve the above equation for $x$ without thinking much, one would definitely raise both sides of the equation to the third power to get rid of the cube root:

$x+a^3=\sqrt[3]{a-x}$

$(x+a^3)^3=(\sqrt[3]{a-x})^3$

$x^3+3x^2a^3+3xa^6+a^9=a-x$

Isolate all terms to the left side, and position them in descending order of $x$ yields

$x^3+3x^2a^3+x(3a^6+1)+a^9-a=0$

This doesn't look great! It's already difficult to solve for the cubic equation for the known coefficients of $x^2,\,x$ and constant, it's doubly hard to solve for cubic equation above, with the parameter $a$ in it.

Okay, we have to stop continuing and start to think from other perspective.

Looking back to the given equality, $x+a^3=\sqrt[3]{a-x}$, what have you noticed?

If you are an expert, then you can already see the "pattern", but what if you are novice to this kind of equation?

Worry not, remember we have all the superb state-of-the art technology that could help us to investigate the possible pattern that might lie in any given mathematical problem!

Let's first use the wolfram alpha to plot for the graph of the function from the LHS and RHS separately, and see where that leads us...

Graph_1

Ops, that doesn't reveal much with the parameter $a$ in there, we could replace $a$ by any figure, just so you know to investigate further:

Let's replace $a$ by $1$:

Graph_2

Aww...that doesn't give us anything valuable there, you might want to try $a=2$ instead.

Graph_3

That doesn't help either, the graph just doesn't help to let ideas sprang to mind...so, giving up seems appropriate!

Nope, you don't do that!

What if you treat $x$ the parameter and $a$ the variable? Aha! Does this idea ever cross your mind? :D

So, now, you replace $x=1$ and $a=x$ into the equations and plot them in the same graph with wolfram:

Graph_4

The two graphs don't look like they have something to tell us...:(

Should we continue to use wolfram for graphing help, or should we try other ICT tool, such as the Desmos Graphing Calculator?

For any graphing purposes, I would like to propose students to single out Desmos Graphing Calculator for the reason it allows us to plot functions, create tables, add sliders, animate our graphs, scale the axes independently, touch a curve to show maximums, minimums, and points of intersection, tap the gray points of interest to see their coordinates, hold and drag along a curve to see the coordinates change under your finger. and more -- all for free.

Now we plot the graphs $y=1+x^3$ and $y=\sqrt[3]{x-1}$ with Desmos Graphing Calculator and we see that we have

Aww...the graphs are like symmetry w.r.t the $y=x$, that reveals a fairly important pattern at this juncture, that we know most probably they are the inverses of each other! We need to ascertain if that is the case!

Let $f(a)=x+a^3$, thus $f(f^{-1}(a))=a$ (where $x$ is parameter) gives us

$x+(f^{-1}(a))^3=a$

$(f^{-1}(a))^3=a-x$

$f^{-1}(a)=\sqrt[3]{a-x}$

Yipee!!! Those two $(f(a)=x+a^3$ and $f^{-1}(a)=\sqrt[3]{a-x})$ are really inverses of each other!

From the given equality where $x+a^3=\sqrt[3]{a-x}$, this only possible if

$f(a)=x+a^3=a$ or $f^{-1}(a)=\sqrt[3]{a-x}=a$

Solving either one gives $x=a-a^3$.

We have not done yet, we need to find the other two roots from the given cubic equation.

Performing polynomial long division to $x^3+3x^2a^3+x(3a^6+1)+a^9-a=0$ by dividing it with $x-a+a^3$, we obtain:

$\small x^3+3a^3x^2+(3a^6+1)x+a(a^8-1)=(x-a+a^3)(x^2+(2a^3+a)x+a^4+a^4+a^2+1)$

Solving the quadratic factor gives us the complex roots

$x=\dfrac{-(2a^3+a)\pm\sqrt{-3a^2-4}}{2}$.