First slide

Theory of equations

Question

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

Moderate

A

$a = - b = - c$
B

$a = 2 b = 2 c$
C

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

None of these

Solution

As $f (x) = x^{3} - 3 b^{2} x + 2 c^{3}$ is divisible by $x - a$ and $x - b$ , therefore
$f (a) = 0 \Rightarrow a^{3} - 3 b^{2} a + 2 c^{3} = 0 . . . (i)$
and $f (b) = 0 \Rightarrow b^{3} - 3 b^{3} + 2 c^{3} = 0 . . . (ii)$
From $(ii), b = c$
From $(i), a^{3} - 3 {ab}^{2} + 2 b^{3} = 0 (putting b = c)$
$\Rightarrow (a - b) (a^{2} + ab - 2 b^{2}) = 0$
$\Rightarrow a = b or a^{2} + ab = 2 b^{2}$
Thus $a = b = c or a^{2} + ab = 2 b^{2} and b = c$
$a^{2} + ab = 2 b^{2}$ is satisfied by $a = - 2 b . But b = c .$
$∴ a^{2} + ab - 2 b^{2}$
and $b = c$ is equivalent to $a = - 2 b = - 2 c$

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