Question

If x + y + z = 0, prove that `(x + y)^2/(xy) + (y + z)^2/(yz) + (z + x)^2/(zx) = 3`

Answer 1

Given: x + y + z = 0

So,

x + y = −z

y + z = −x

z + x = −y

Let’s substitute it into the expression:

`(x + y)^2/(xy) + (y + z)^2/(yz) + (z + x)^2/(zx)`

And, it becomes,

`(-z)^2/(xy) + (-x)^2/(yz) + (-y)^2/(zx)`

= `z^2/(xy) + x^2/(yz) + y^2/(zx)`

Rewriting the expression:

∴ `x^2/(xy) + y^2/(yz) + z^2/(zx)`

Taking LCM of the denominators:

LCM of yz, zx, xy is xyz,

= `(x^2(x) + y^2(y) + z^2(z))/(xyz)`

= `(x^3 + y^3 + z^3)/(xyz)`

Using the identity:

When x + y + z = 0, then:

x³ + y³ + z³ = 3xyz    ...(As per identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc)

Hence,

`(x^3 + y^3 + z^3)/(xyz) = (3xyz)/(xyz)`

∴ `(x^3 + y^3 + z^3)/(xyz) = 3`

∴ `(x + y)^2/(xy) + (y + z)^2/(yz) + (z + x)^2/(zx) = 3`