Question

If a, b and c are real and $x^3-3b^{2}x+2c^3$ is divisible by $(x-a)$ and $(x-b),$ then

• A. $a=-b=-c$
• B. $a=2b=2c$
• C. $4b^2=ac$
• D. None of these

Answer 1

Answer: (D)

As $f(x)=x^3-3b^{2}x+2c^2$ is divisible by $(x-a)$ and $(x-b)$ . Now, $f(a)=0$ $\Rightarrow$ $a^3-3b^{2}a+2c^3=0$ ...(i) and $f(b)=0$ $\Rightarrow$ $b^3-3b^3+2c^3=0$ ...(ii) From Eq. (ii), $b=c$ On putting $b=c$ in Eq. (i), we get $a^3-3a{b}^2+2b^3=0$ $\Rightarrow$ $(a-b)(a^2+ab-2b^2\}=0$ $\Rightarrow$ $a=b$ or $a^2+ab=2b^2$ Thus, $a=b=c$ or $a^2+ab=2b^2$ and $b=c$ Therefore, $a^2+ab=2b^2$ and $b=c$ is equivalent to $a=-2b=-2c$