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I know that I can see that this match with $y = 0$, $x = 0$ and $y = -x$.

But, the author of the book says:

"Hint: From the assumption $(x + y)^5 = x^5 + y^5$ you should be able to derive the equation $x^3 + 2x^2y + 2xy^2 + y^3 = 0$ if $xy \neq 0$. This implies that $(x + y)^3 = x^2y + y^2x = xy(x + y)$"

Then, I realize that $(x + y)^3 - xy(x + y) = x^3 + 2x^2y + 2xy^2 + y^3$.

That's why if $(x + y)^3 - xy(x + y) = 0$ then $(x + y)^3 = xy(x + y)$.

Then, I did the same with $(x + y)^5$

$x^5 + 5x^4y +9x^3y^2 + 9x^2y^3 + 5xy^4 + y^5 = 0$ if $xy \neq 0$

Then $(x + y)^5 - x^2y^2(x + y) = 0$

$(x + y)^5 = x^2y^2(x + y)$ Then I can prove that its true when $y = -x$

But I still don't get it why I have to did this and what the author want to tell me doing this.

Sil
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Jonatan Garcia
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1 Answers1

6

Hint: From the assumption $(x + y)⁵ = x⁵ + y⁵$ you should be able to derive the equation $x³ + 2x²y + 2xy² + y³ = 0$ if $xy \neq 0$.

Using the binomial expansion:

$$\require{cancel} \begin{align} (x+y)^5=x^5+y^5 \;\;&\iff\;\; \cancel{x^5} + 5x^4y +10x^3y^2 + 10x^2y^3 + 5xy^4 + \bcancel{y^5} = \cancel{x^5}+\bcancel{y^5} \\ &\iff\;\; 5xy\,(x^3+2x^2y+2xy^2+y^3) = 0 \end{align} $$

Assuming that $\,xy\ne 0\,$, this implies $\,x^3+2x^2y+2xy^2+y^3 = 0\,$

But I still don't get it  [...]  what the author want to tell me doing this

Note that $\,x^3+2x^2y+2xy^2+y^3 = (x + y) (x^2 + x y + y^2)\,$, and the only (real) zeros occur for $\,x+y=0\,$ since the quadratic factor is always strictly positive if $\,xy \ne 0\,$.

dxiv
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