Question

The roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$ are equal then

• A. $a^3+b^3+c^3=3abc$ or $a=0$
• B. $a+b+c=0$
• C. $a^3+b^3+c^3=3abc$ or $a=1$
• D. none of these

Answer 1

The correct option is A $a^3+b^3+c^3=3abc$ or $a=0$

$4(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$

root are equal so D = 0

$4{(a^2-bc)}^2-4(c^2-ab)(b^2-ac)=0a^4-2a^{2}bc+b^2{c}^2-(b^2{c}^2-{ab}^3-a{c}^3=0a(a^3+b^3+c^3-3abc)=0\therefore a=0or{a}^3+b^3+c^3=3abc$