What are the possible values of 1/a + 1/b when a^3+3a^2b^2+b^3 = a^3b^3?

[anemone]

Here is this week's POTW:

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Let $a$ and $b$ be non-zero real numbers such that $a^3+3a^2b^2+b^3=a^3b^3$.

Find all the possible values of $\dfrac{1}{a}+\dfrac{1}{b}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

Hello, MHB Community! (Wave)

anemone has asked me to fill in for her while she is away on holiday. :D

Congratulations to to following users for their correct submissions:

• kaliprasad
• castor28

Honorable mention goes to lfdahl for a partially correct solution.

castor28's solution follows:

Spoiler

By dividing by $a^3b^3$, the original equation can be written as:
$$\frac{1}{a^3}+\frac{3}{ab}+\frac{1}{b^3} = 1$$

On the other hand, with $u=\dfrac1a+\dfrac1b$, we have:
$$\begin{align*}
u^3 &= \frac{1}{a^3} + \frac{3}{a^2b} + \frac{3}{ab^2} + \frac{1}{b^3}\\
&= \frac{1}{a^3} + \frac{3u}{ab}+ \frac{1}{b^3}
\end{align*}$$

Subtracting, we get:
$$u^3-1=\frac{3(u-1)}{ab}$$

This already shows that $u=1$ is a solution. If $u\ne1$, we have:
$$\begin{align*}
u^2 + u + 1 &= \frac{3}{ab}\\
\frac{1}{a^2} + \frac{2}{ab} + \frac{1}{b^2} + \frac{1}{a} + \frac{1}{b} + 1 &= \frac{3}{ab}\\
\frac{1}{a^2} - \frac{1}{ab} + \frac{1}{b^2} + \frac{1}{a} + \frac{1}{b} + 1 &= 0\\
x^2 - xy + y^2 + x + y + 1 &=0
\end{align*}$$

with $x=\dfrac1a$ and $y=\dfrac1b$. Multiplying by 4 and completing the squares, the last equation gives:
$$(2x-y+1)^2 + 3(y+1)^2 = 0$$

Since both squares are non-negative, the only real solution of that equation is $y=-1$, $x=-1$, giving $u=x+y=-2$.

To summarize, the only possible values of $u=\dfrac1a+\dfrac1b$ are $1$ and $-2$.