If x, y, z are any three non-zero real numbers such that x4 - y2 = z2, y4 - z2 = x2 and z4 - x2 = y2 then (x2 + 1)-1 + (y2 + 1)-1 + (z2 + 1)-1 =

Given:

x⁴ - y² = z²

y⁴ - z² = x²

z⁴ - x² = y²

We need to find the value of:

(x² + 1)^-1 + (y² + 1)^-1 + (z² + 1)^-1

Formula used:

The given equations are:

1. x⁴ - y² = z²

2. y⁴ - z² = x²

3. z⁴ - x² = y²

Calculations:

We are given the symmetric nature of the equations. Let’s assume that:

x = y = z

Substituting this assumption into the first equation:

x⁴ - x² = x²

⇒ x⁴ = 2x²

⇒ x²(x² - 2) = 0

Since x ≠ 0, we have:

x² = 2

⇒ x = y = z = √2

Now, substitute x = y = z = √2 into the given expression:

(x² + 1)^-1 + (y² + 1)^-1 + (z² + 1)^-1

⇒ 3 × (2 + 1)^-1

⇒ 3 × 3^-1

⇒ 3 × (1/3)

⇒ 1

∴ The value of the expression is 1.