Question

Can this be done without trigonometry?

Answer 1

If x+y+z=xyz. Prove that
\[\frac{ 2x }{ 1-x ^{2} }+\frac{ 2y }{ 1-y ^{2} }+\frac{ 2z }{ 1-z ^{2} }=\frac{ 2x }{ 1-x ^{2} }*\frac{ 2y }{ 1-y ^{2} }*\frac{ 2z }{ 1-z ^{2} }\]

Answer 2

@Kainui @dan815 @ganeshie8

Answer 3

take lcm @No.name

Answer 4

2x(1-y^2)(1-z^2)+2y(1-x^2)(1-z^2)+2z(1-x^2)(1-y^2)/((1-x^2)(1-y^2)(1-z^2))

Answer 5

x+y+z-xz^2-y^2x-z^2y-x^2y-zy^2-zx^2+y^2z^2x+x^2z^2y+z*x^2*y^2/ denominator

Answer 6

(x+y+z)^3 format

Answer 7

(a+b+c)^3= a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^c + 3ac^2 + 3b^2c + 3bc^2 + 6abc

Answer 8

I sort of got stuck part way through, but I think this is getting pretty close to what you want. Hopefully someone can figure out the rest.

I noticed that we have: \[\Large \frac{8xyz}{(1-x^2)(1-y^2)(1-z^2)}\] which given x+y+z=xyz means: \[\Large \frac{8(x+y+z)}{(1-x^2)(1-y^2)(1-z^2)}=\]\[4\frac{2x}{(1-x^2)(1-y^2)(1-z^2)}+4\frac{2y}{(1-x^2)(1-y^2)(1-z^2)}+4\frac{2z}{(1-x^2)(1-y^2)(1-z^2)}\] So if you can show that \[\LARGE \frac{4}{(1-x^2)(1-y^2)}=1\] then since x, y, and z are symmetric, you will be done.

Perhaps you should consider \[\LARGE x^2y^2z^2=x^2+y^2+z^2+2xy+2yz+2xz\]

Answer 9

Oh i got what u r trying to say