<s>[INST] roots of the equation $(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$ are rational $a,b,c$ are non zero , unequal rational numbers then prove that roots of the equation $$(abc^2)x^2+3a^2 cx+b^2cx-6a^2-ab+2b^2=0$$ are rational
quadratic eqn. in standard form $(abc^2)x^2+(3a^2c+b^2c)x-(6a^2+ab-2b^2) = 0$
$\displaystyle D=(3a^2c+b^2c)^2+4(abc^2)(6a^2+ab-2b^2) = 9a^4c^2+b^4c^2+6a^2b^2c^2+24a^3bc^2+4a^2b^2c^2-8ab^3c^2$
could some help me with this
 [/INST] $$B=3a^2c+b^2c=c(3a^2+b^2)$$
$$-C=6a^2+ab-2b^2=6a^2+4ab-3ab-2b^2=2a(3a+2b)-b(3a+2b)=(3a+2b)(2a-b)$$
$$A=abc^2$$
Break $$C\cdot A=(3a+2b)ac\cdot bc(2a-b)$$  as $$(3a+2b)ac-bc(2a-b)=B$$
</s>