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Trying to find when: $x^3+2x^2y+2xy^2+y^3=0$

The solution in my book makes the observation: $(x+y)^3=x^3+3x^2y+3xy^2+y^3$

This is similar to our equation with the difference being: $x^2y+xy^2$

Simplify to: $xy(x+y)$, and we see that this is equal to zero for $x=0, y=0, x=-y$.

I don't understand how we transitioned our question from the original equation to a question about this final equation vis a vis taking that difference in the middle step.

lopan
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2 Answers2

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Addendum added to respond to the comment/question of lopan.

The book apparently left out some details.

$(x + y) = 0 \iff x = -y$.

Therefore, you can assume, without loss of generality, that $(x + y) \neq 0$, and then explore what happens, based on this assumption.

From the constraints, you have that

$$(x+y)^3 - (xy)(x+y) = 0. \tag1 $$

Since it is being assumed that $(x+y) \neq 0$, you can divide both sides of (1) above by $(x+y)$.

This yields:

$$(x+y)^2 - xy = 0 \implies x^2 + xy + y^2 = 0. \tag2 $$

There are a number of semi-sophisticated ways to proceed from here. The least sophisticated (i.e. simplest) is to construe the equation in (2) above as a quadratic equation in $x$.

Then

$$x = \frac{1}{2} \left[-y \pm \sqrt{y^2 - 4y^2}\right]$$

$$= \frac{1}{2} \left[-y \pm \sqrt{-3y^2}\right]. \tag3 $$

The only way that $x$ can be set to a real value is if the expression $(-3y^2)$ is non-negative.

This requires that $y = 0$ which then implies that $x = 0.$

Addendum

Responding to the comment question of lopan:

But I don't see how the whole thing starts, why is the question about when $(x+y)=0$ and not about when the original equation is zero?

Let $f(x,y) = x^3+2x^2y+2xy^2+y^3.$

The original problem is to find all (presumably real numbers) $(x,y)$ such that $f(x,y) = 0.$

As you indicated at the start of your analyis,

$$f(x,y) = (x + y)^3 - (x+y)(xy).$$

Therefore, the original problem can be re-stated as:

$$(x+y) \times \left[(x+y)^2 - xy\right] = 0. \tag3 $$

Therefore, if $(x + y) \neq 0$ then you must have that

$$(x+y)^2 - xy = 0.$$

user2661923
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  • But I don't see how the whole thing starts, why is the question about when $(x+y)=0$ and not about when the original equation is zero? – lopan Dec 04 '22 at 02:41
  • A relatively simple way is to note that $(x+y)^2 - xy = 0 ; \to ; (x+y)^2 = xy$. Since $x + y \neq 0$, we have $(x+y)^2 \gt 0 ; \to ; xy \gt 0$. Also, $x^2 \ge 0$ and $y^2 \ge 0$, so $x^2 + xy + y^2 \gt 0$. – John Omielan Dec 04 '22 at 02:50
  • @JohnOmielan Yes, but in my (highly subjective) opinion, for the beginning Math student, a more useful generic weapon in such situations is to treat the equation as a quadratic in $(x)$. – user2661923 Dec 04 '22 at 02:52
  • Thanks, I will need some time to try to digest this. – lopan Dec 04 '22 at 02:54
  • @lopan See the Addendum that I have just added to the end of my answer. – user2661923 Dec 04 '22 at 02:56
  • Ah, now I see the missing link. Essentially we are completing the square (I guess here the "cube"). Then we have the junk left over on the right side of the equals, and we simply move it back over to the left by subtracting from right. So, we're still working with the equivalent of the original equation since we essentially added the new term to both sides, but the one on the left was incorporated into the $(x+y)^3$. – lopan Dec 04 '22 at 03:10
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if $y \neq 0,$ define $r = \frac{x}{y},$ divide the original equation by $y^3$ to get $$ r^3 + 2 r^2 + 2r + 1 = 0$$ This factors, $$ (r+1) (r^2 + r + 1) = 0$$

in the OP it mentions $r=-1,$ and this is the only real ratio. If you are allowing complex numbers, the other two solutions are $$ r = \frac{-1 + i \sqrt 3}{2} \; , \; \; \; r = \frac{-1 - i \sqrt 3}{2} $$ which are cube roots of $1$

Will Jagy
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