Show that the roots of the equation

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always real and they cannot be unless a = b = c.

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0

⇒ x(x – b) – a(x – b) + x(x – c) – b(x – c) + x(x – a) – c(x – a) = 0

⇒ x² – bx – ax + ab + x² – cx – bx + bc + x² – ax – cx + ac = 0

⇒ 3x² – 2x(a + b + c) + ab + bc + ac = 0

D = b² – 4ac

D = (a + b + c)² – 4(3)(ab + bc + ac) = 0

D = 4(a² + b² + c² + 2ab + 2bc + 2ca – 3ab – 3bc – 3ca)

D = 4(a² + b² + c² – ab – bc – ca)

D = 2[(a –b)² + (b – c)² + (c – a)²]

Which is always greater than zero so the roots are real.

Roots are equal if D = 0

i.e. (a – b)² + (b + c)² + (c – a)² = 0

since sum of three perfect square is equal to zero so each of them separately equal to zero.

So, a – b = 0, b – c = 0, c – a = 0

a = b , b = c, c = a

so, a = b = c.