Q.If $a+b+c=0$ and $a, b, c$ are rational numbers then, prove that the roots of the equation $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ are rational numbers.

Rewrites each bracket using $a+b+c=0$, reducing the cubic-looking coefficients to $-2a,-2b,-2c$, then shows the resulting quadratic $ax^2+bx+c=0$ has a perfect-square discriminant.

1. Since $a+b+c=0$: $b+c=-a\Rightarrow b+c-a=-2a$; similarly $c+a=-b\Rightarrow c+a-b=-2b$; and $a+b=-c\Rightarrow a+b-c=-2c$.
2. Substituting, the equation $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ becomes $-2ax^2-2bx-2c=0$, i.e. (dividing by $-2$, assuming $a\ne0$) $ax^2+bx+c=0$.
3. Discriminant: $D=b^2-4ac$. Using $c=-(a+b)$: $D=b^2-4a(-(a+b))=b^2+4a^2+4ab=4a^2+4ab+b^2=(2a+b)^2$. …