Show that the roots of (x-b)(x-c) +(x-c)(x-a) +(x-a)(x-b) =0 are real, and that they cannot be equal unless a=b=c.

(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b)

= x² - bx - cx + bc + x² - cx - ax + ca + x² - ax - bx + an

= x² + x² + x² - ax - ax - bx - bx - cx - cx + ab + bc + ca (Rearranging the equation)

= 3x² - 2ax - 2bx - 2cx + (ab + bc + ca)

= 3x² - 2x(a + b + c) + (ab + bc +ca)

= 3x² - 2(a + b + c)x + (ab + bc + ca)

Now discriminant for above equation

D = { - 2(a + b + c)}² - 4 x 3 ( ab + bc + ca)

= 4(a + b + c)² - 12(ab + bc + ca)

= 4 [ (a + b + c)² - 3(ab + bc + ca)]

= 2 x 2 x [ a² + b² + c² + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca]

= 2 x 2 x [ a² + b² +c² - ab - bc -ca]

= 2 [ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca] ( Multiplying the quantity in bracket by 2)

= 2 [ (a² + b² - 2ab) + (b² + c² - 2bc) + (c² + a² - 2ca) ] ( Rearranging the equation)

= 2 [ (a - b)² + (b - c)² + (c - a)²]

Now since square of any quantity is either 0 or greater than zero that is it's always positive.

= (a - b)² , (b - c)² (c - a)² are all positive numbers.

Thus there sum is also a positive number

Thus the double of their sum that is

2 [ (a - b)² + (b - c)² + (c - a)²] is also positive.

Now since discriminant is positive that is D = 0

= Roots are real.

Second part :

For roots to be equal discriminant D = 0

= 2 [ (a - b)² + (b - c)² + (c - a)²] = 0

= (a - b)² + (b - c)² + (c - a)² = 0

Since square numbers are always positive thus the above sum can be zero only if

a - b = 0, b - c = 0 c - a = 0

= a = b, b = c , c = a

= a = b = c