First slide

Theory of equations

Question

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

Moderate

A

$a = - b = - c$
B

$a = 2 b = 2 c$
C

$a = b = c$ or $a = - 2 b = - 2 c$
D

none of these

Solution

It is given that $f (x) = x^{3} - 3 b^{2} x + 2 c^{3}$ is divisible by $x - a$ and $x - b .$
$∴ f (a) = 0$ and $f (b) = 0$
$\Rightarrow a^{3} - 3 b^{2} a + 2 c^{3} = 0$ …(i)
and $b^{3} - 3 b^{3} + 2 c^{3} = 0$ …(ii)
From (ii), we get $b = c .$
Putting, $b = c$ in (i), we get
$\begin{array}{l} a^{3} - 3 a b^{2} + 2 b^{3} - 0 \\ \Rightarrow (a - b) (a^{2} + a b - 2 b^{2}) = 0 \end{array}$
$\Rightarrow a = b$ or $\Rightarrow a = b$
Thus, $a = b = c$ or $, a^{2} + a b = 2 b^{2}$ and $b = c$
Clearly, $a^{2} + a b = 2 b^{2}$ is satisfied by $a = - 2 b .$
$a^{2} + a b = 2 b^{2}$ and $b = c$
$\Rightarrow a = - 2 b$ and $b = c \Rightarrow a = - 2 b = - 2 c$

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