Solve by using only & only …

The given equation is; (x - a)(x - b)+(x - b)(x - c)+(x - c)(x - a) =0 
{tex}\Rightarrow{/tex} (x² - ax - bx + ab) + (x²- bx - cx + bc) +(x² - cx - ax + ac) = 0
{tex}\Rightarrow{/tex}3x² - 2x(a + b + c) + (ab + bc + ca) = 0.....(1). 
Discriminant 'D' of quadratic equation (1) is given by;
{tex}\therefore{/tex} D = 4(a + b + c)² - 12(ab + bc + ca)
= 4[(a + b + c)² - 3(ab + bc + ca)]
= 4(a² + b² + c² - ab - bc - ca)
= 2(2a² + 2b² + 2c² - 2ab - 2bc - 2ca)
= 2 [(a - b)² + (b - c)² + (c - a)²] ≥ 0
[{tex}\because{/tex} (a - b)² ≥ 0, (b - c)² {tex}{/tex} ≥ 0 and (c - a)² {tex}{/tex} ≥ 0].
This shows that both the roots of the given equation are real.
For equal roots, we must have D = 0.
Now, D = 0 {tex}\Rightarrow{/tex} (a - b)² + (b - c)² + (c - a)²= 0
{tex}\Rightarrow{/tex} (a - b) = 0, (b - c) = 0 and (c - a) = 0 (sum of squares can be zero only if they all are equal to 0)
{tex}\Rightarrow{/tex} a = b = c.
Hence, the roots are equal only when a = b = c