Question

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

• A. a = b = c, a = -2b = -2c
• B. a = -b = -c
• C. a = 2b = 2c
• D. a=b=c , a=2b=2c

Answer 1

The correct option is A

a = b = c, a = -2b = -2c

As $f\left(x\right)=x^3-3b^{2}x+2c^3$ is divisible by $x-a$ and $x-b$, therefore

$f\left(a\right)=0$ ⇒ $a^3-3b^{2}a+2c^3=0$ ......(i)

and f(b) = 0 ⇒ $b^3-3b^3+2c^3=0$ .... (ii)

From (ii), b = c

From (i), $a^3-3a{b}^2+2b^3=0$ (putting b = c)

⇒ $(a-b)(a^2+ab-2b^2)=0$

⇒ a = b = c or $a^2+ab=2b^2$ and b = c

∴ $a^2+ab-2b^2$

and b = c is equivalent to a= -2b = -2c