But I Don't Know The Exact Values!

Let us first expand the terms on the LHS to get a better idea of the coefficients of the quadratic equation.

$\begin{aligned} & \color{#D61F06}{x^2 - (a +b)x + ab} + \color{#20A900}{x^2 - (b +c)x + bc} + \color{#3D99F6}{x^2 - (c +a)x + ca} = 0 \\ \iff & 3x^2 - 2 (a + b + c) x + (ab + bc + ca) = 0 \end{aligned}$

If its roots of this polynomial are equal, then the discriminant of this quadratic equation is equal to zero.

$\begin{aligned} & \left(-2(a + b + c) \right)^2 - 4 \times 3 \times (ab + bc + ca ) = 0 \\ \iff & 4(a + b + c)^2 - 4 \times 3 \times (ab + bc + ca ) = 0 \\ \iff & (a + b + c) ^2 - 3 \times (ab + bc + ca ) = 0 \\ \iff & a^2 + b^2 + c^2 - (ab + bc + ca) = 0 \\ \end{aligned}$

The LHS can be written as $\frac{1}{2} \left(\color{#D61F06}{a^2 + b^2 - 2ab} + \color{#20A900}{b^2 + c^2 - 2bc} + \color{#3D99F6}{c^2 + a^2 - 2ca} \right) = 0$

This terms on the LHS can be factorized as $\frac{1}{2} \left( \color{#D61F06}{(a-b)^2} + \color{#20A900}{(b-c)^2} + \color{#3D99F6}{(c- a)^2} \right) = 0$ .

By the trivial inequality , any perfect square is always non-negative. Thus for the LHS to be zero, each of the perfect square should be zero at the same time. This is only possible when $a =b =c$ . $_\square$