Solving the following equations : $yz=a\left( y+z \right) +\alpha ,\ zx=a\left( z+x \right) +\beta ,\ xy=a\left( x+y \right) +\gamma $ we get $x=a\pm \sqrt { \left{ \dfrac { \left( { a }^{ 2 }+\beta \right) \left( { a }^{ 2 }+\gamma \right) }{ { a }^{ 2 }+\alpha } \right} }$

Add y to both sides of the first equation and take out y as a common factor to get y(z-a) = az + alpha, which means y = (az + alpha) / (z-a). Applying the exact same algebraic steps to the other two equations yields x = (az + beta) / (z-a) and x = (ay + gamma) / (y-a). By substituting the expression for y into the equation for x, then applying that result into the equation for z, and finally using the original equation for x, you solve for z and then x. This extensive algebraic substitution and simplification of the rational expressions confirms that the derived value for x is exactly a plus or minus the square root of the product of (a^2 + beta) and (a^2 + gamma) divided by (a^2 + alpha). The statement is true.

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