Another Tricky FE

[Nirjhor]

Find all \(f:\mathbb{R}\mapsto\mathbb{R}\) satisfying \[(x-y)f(x+y)-(x+y)f(x-y)=4xy\left(x^2-y^2\right)\] for all real \(x,y\).

Try this.

[SANZEED]

Solution:

Let $f(x)=x^{3}+g(x) \forall x\in \mathbb{R}$. Now the given statement can be written as $(x-y)(x+y)^{3}+(x-y)g(x+y)-(x+y)(x-y)^{3}-(x+y)g(x-y)=4xy(x^{2}-y^{2})$. Simplification yields $(x-y)g(x+y)=(x+y)g(x-y)$. Let us denote the last statement by $Q(x,y)$. Then $Q(x+1,x)\Rightarrow g(2x+1)=(2x+1)g(1)$. Now take $x=\dfrac{y-1}{2}$. Then $g(y)=yg(1)\forall y\in \mathbb{R}$. Thus the solution is $f(x)=x^{3}+cx \forall x\in \mathbb{R}$.

[Nirjhor]

Nice substitution!

My solution: Because \(4xy=(x+y)^2-(x-y)^2\), the equation can be rewritten as \[\dfrac{f(x+y)}{x+y}-(x+y)^2=\dfrac{f(x-y)}{x-y}-(x-y)^2\] so \(f(x)/x - x^2\) is constant. Letting the constant \(c\) and solving leads to \(f(x)=x^3+cx\) for all \(x\).

The similar approach works for the previous Tricky FE I posted.

[*Mahi*]

If you're using $\dfrac{f(x)}x -x^2 =$ constant, you have to prove $f(0)=0$ separately as $g(x) = \dfrac{f(x)}x -x^2$ is not defined at $x=0$.

$x=\frac{a+b}2, y=\frac{a-b}2$ can also be handy whenever the terms are sums and products of $x+y \text{ or } x-y$