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$x,y \in \mathbb{R}\setminus\{0\}$

Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$ But it’s the same thing.

My Attempt: $$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$ $$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$

But i don’t know what to do now. Please if you know the key to this type of problems ‘Find all possible values’ post it.

PNT
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4 Answers4

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Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have $$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3} =1-\frac{3}{xy} \tag{1}$$ But we have also

$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$ From (1) and (2), we deduce \begin{align} &(\frac{1}{x}+\frac{1}{y})^3 = (1-\frac{3}{xy}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \\ &\iff (\frac{1}{x}+\frac{1}{y})^3- 1=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1 \right) \\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1 \right)=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1 \right) \\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1 -\frac{3}{xy} \right)=0\\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} +(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy} \right)=0\\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}-\frac{1}{y})^2 +\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( ((1+\frac{1}{x})-(1+\frac{1}{y}))^2 +\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\ &\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( \left(1+\frac{1}{x} \right)^2+\left(1+\frac{1}{y} \right)^2 -\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\ \end{align}

The last equation happens if and only if $\frac{1}{x}+\frac{1}{y} =1$ or $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $. Hence, the two possible values of $\frac{1}{x}+\frac{1}{y}$ is $1$ and $-2$.

NN2
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  • How did you move from the first equation to the second – PNT Feb 08 '21 at 10:32
  • You can replace $\frac{1}{x^3}+\frac{1}{y^3} $ by $1-\frac{3}{xy}$ in the second equation – NN2 Feb 08 '21 at 10:36
  • Is there a mis-step in saying $(\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1-\frac{3}{xy}=((\frac{1}{x}-\frac{1}{y})^2 +(1+\frac{1}{x})(1+\frac{1}{y}))$. I seem to be missing something. – WA Don Feb 08 '21 at 10:36
  • @WADon: no, you have: $(\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} = (\frac{1}{x}-\frac{1}{y})^2$ and $(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy} = (1+\frac{1}{x})(1+\frac{1}{y})$ – NN2 Feb 08 '21 at 10:38
  • Oh yes, thanks! – WA Don Feb 08 '21 at 10:41
  • If $\frac{1}{x}+\frac{1}{y}=-2$ then $y=-\frac{x}{2 x+1}$ and $$x^3+3 x^2 y^2+y^3=x^3 y^3$$ is not satisfied – Raffaele Feb 08 '21 at 11:21
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    @Raffaele: No, $x=y=-1$ (because $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $). So, we deduce $\frac{1}{x}+\frac{1}{y} = -2$. But you don't have $y = -\frac{x}{2x+1}$ for all $x$ here – NN2 Feb 08 '21 at 12:00
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    @NN2: One can shorten the calculation a bit by introducing $s=1/x+1/y$ and $p=1/(xy)$. Then $0=s^3 -3ps+3p-1=(s-1)(s^2+s-3p+1)$. The second factor is $\ge s^2+s-3p/4+1 = (s+2)^2/4$ because of AM-GM, so it becomes zero only for $s=-2$. – Martin R Feb 08 '21 at 12:05
  • @MartinR: I agree. – NN2 Feb 08 '21 at 12:07
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Hint:

Let $\dfrac{x+y}{xy}=k$

$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$

$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$

$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$

As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$

Clearly, $k=1$ is a solution.

Otherwise, $$xy=\dfrac3{1+k+k^2}$$

lab bhattacharjee
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1

$$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$ $$\frac{x+y}{xy}\tag{2}$$ $$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$ can be factorized$^1$ as $$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$ $x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are only those of $x+y-xy=0$ that is $x+y=xy$ and finally $$\frac{x+y}{xy}=1$$

$(^1)$

$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=(x+y)^3-3x^2y-3xy^2+ 3 x^2 y^2 - x^3 y^3=$$ $$=\left[(x+y)^3- x^3 y^3\right]+\left[-3x^2y-3xy^2+ 3 x^2 y^2 \right]=$$ $$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)\right]-3 x y (x+y-x y)=$$ $$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)-3xy\right]=(x+y-xy)\left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)$$

$(^2)$

$$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$$ set $y=kx$. We get $$k^2 x^4+k^2 x^3+k^2 x^2+k x^3-k x^2+x^2=0\to x^2 \left(k^2 x^2+k^2 x+k x+k^2-k+1\right)=0$$ As $x\ne 0$ we have $$k^2 x^2+k^2 x+k x+k^2-k+1=0$$ the discriminant of this equation is $D=-k^4+2 k^3-k^2=-k^2(k-1)^2$

$D\ge 0\to k=1 $ (solution $k=0$ is discarded)

if $k=1$ we have $x^2+2x+1=0\to x=-1$ thus the solutions are $$x=-1;\;y=-1$$

Raffaele
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The following is equivalent to @NN2's solution, only written a bit differently.

First substitute $x=1/a, y=1/b$, so that the equation becomes $$ \tag{*} a^3+b^3+3ab - 1 = 0 \, . $$ We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then $$ 0 = a^3+b^3+3ab - 1 = S^3+3PS + 3P -1 = (S-1)(S^2 + S-3P+1) \, . $$

So $S=1$ is one possible value for $a+b$. Conversely, all $(a, b)$ with $a+b=1$ satisfy the equation $(*)$.

In order to determine the possible zeros of the second factor we use that $P \le S^2/4$ from the inequality between geometric and arithmethic mean: $$ 0 = S^2 + S-3P+1 \ge S^2 + S-\frac 34 S^2+1 = \frac 14 (S+2)^2 \ge 0 $$ is only possible if $S=-2$ and $P=1$, i.e. if $(a, b) = (-1, -1)$.

Therefore $S=a+b=1/x+1/y$ can only take the values $1$ and $-2$, and both values do actually occur.

Martin R
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