Suppose that \(\displaystyle \alpha\in\mathbb{R}\setminus\{0\}\), surely if \(\displaystyle \bf x=y=z=\alpha\) works?Determine all three (x, y, z) real numbers other than 0 for which the equation is met xy(x+y)=yz(y+z)=zx(z+x)
Thank you for your help but how to prove it? How do you demonstrate this by turning into equations.Suppose that \(\displaystyle \alpha\in\mathbb{R}\setminus\{0\}\), surely if \(\displaystyle \bf x=y=z=\alpha\) works?
Determine all three (x, y, z) real numbers other than 0 for which the equation is met xy(x+y)=yz(y+z)=zx(z+x)
I have calculated that for sure x = y = z = 2 but I can't write it mathematically.
Suppose that \(\displaystyle \alpha\in\mathbb{R}\setminus\{0\}\), surely if \(\displaystyle \bf x=y=z=\alpha\) works?
If \(\displaystyle x=y=z=\alpha\ne 0\) then \(\displaystyle xy(x+y)=yz(y+z)=zx(z+x)=2\alpha^3\).Thank you for your help but how to prove it? How do you demonstrate this by turning into equations.