Question

$\begin{array}{c}\text{If a, b, c are distinct real numbers}\text{and}{a}^3+b^3+c^3=3abc,\text{then the equation}\\ a{x}^2+bx+c=0\text{has two roots, out of which one}\text{root is}\end{array}$

• A. $\frac{b}{a}$
• B. $\frac{c}{a}$
• C. $\frac{-b}{a}$
• D. $0$

Answer 1

The correct option is B $\frac{c}{a}$

Given: $a^3+b^3+c^3=3abc$ where $a,b,c$

are distinct real number.

So, $a+b+c=0$

$b=-(a+c)-\text{(i)}$

given that,

$\begin{array}{c}a{x}^2+bx+c=0\\ \text{from (i)}\\ a{x}^2-(a+c)x+c=0\\ a{x}^2-ax-cx+c=0\\ x(ax-c)-1(ax-c)=0\\ (x-1)(ax-c)=0\end{array}$

$x=1$

$ax-c=0$

$x=\frac{c}{2}$

option $B$ is correct.