Prove that (x+y)^3 = x^3 + y^3 for only x =0 and y = 0

Question

Hey all the title pretty much says it all! I'm positive that the only solution in this case is 0 but I'm not entirely sure how to go about proving that... I was thinking cases but I don't think that is super helpful...

Answer 1

Suppose $x^3+y^3=(x+y)^3=x^3+3x^2y+3xy^2+y^3$. Then $3xy(x+y)=0$. This is not true for all $x$ and $y$, but only if one of them is $0$, or the negative of the other. Perhaps this is what you were asked to prove.

Edit: The question title is still wrong - its not true to say that this only holds for $x,y=0$. It also holds for $x=-y$.

Answer 2

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

$$(x+y)^3 = x^3 + y^3 \iff 3x^2y + 3x y^2=0$$

$$ 3xy(x+y)=0 $$

Thus if $x=-y$ or $x=0$ or $y=0$ then,

$$(x+y)^3 = x^3 + y^3$$

Answer 3

$$(x+y)^3=x^3+y^3$$ $$x^3+3x^2y+3xy^2+y^3=x^3+y^3$$ Cancelling out: $$3x^2y+3xy^2=0$$ Factorising: $$3xy(x+y)=0$$

From this we can tell that the statement only works when $x=0$, $y=0$ or when $x+y=0$.

Therefore your statement is incorrect as it also works when $x+y=0$.

Answer 4

If $(x +y)^3 = x^3 + y^3$

$x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3$

$3x^2y + 3xy^2 = 0$

$3x^2y + 3xy^2 = 0$.

Procedure 1:

$3x^2y = -3xy^2$

$x^2y = -xy^2$

If $x = 0$ then this equation holds.

If $x \ne 0$ then

$xy = - y^2$

If $y = 0$ then this equation holds.

If $y \ne 0$ then

$x = -y$.

So there are three possible options. $x = 0; y=0;$ or $x = -y$.

Procedure 2:

$3x^2y + 3xy^2 = 0$

$3xy(x +y) = 0$.

$xy(x+y) =0$.

Then either $xy = 0$ or $x+y = 0$.

If $xy = 0$ then either $x=0$ or $y=0$.

And if $x+y = 0$ then $x = -y$.

So there are three options $x=0; y=0;$ or $x=-y$.

Answer 5

For fun:

$(x+y)^3= (x+y)(x+y)^2;$

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

We have:

$(x+y)^3= x^3+y^3$ ;or

$(x+y)(x+y)^2=$

$(x+y)(x^2-xy+y^2)$, or

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

This implies :

1) $x+y=0$ ; or

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

Then:

$3xy=0$; hence:

$x =0$ or $y=0$.

Altogether:

$x+y=0$ or $x=0$ or $y=0$.