If a^2+b^2+c^2=r^2 , then the value of ⁻¹ ( a b c r )+ ⁻¹ ( b c a r )+ ⁻¹ ( c a b r )=

Step-by-step solution

Given the expression S = ⁻¹ ( a b c r ) + ⁻¹ ( b c a r ) + ⁻¹ ( c a b r ) , define x = a b c r , y = b c a r , and z = c a b r . The products xy , yz , and zx simplify as: xy = b^2 r^2 , yz = c^2 r^2 , and zx = a^2 r^2 . Their sum is xy + yz + zx = a^2 + b^2 + c^2 r^2 . Since a^2 + b^2 + c^2 = r^2 , it follows that xy + yz + zx = 1 . Given x, y, z > 0 for positive a, b, c, r , the identity ⁻¹x + ⁻¹y + ⁻¹z = 2 holds. S = 2