The L.C.M. of $9 x^{2} + 6 x + 1, 3 x^{2} + 7 x + 2$ and $2 x^{2} + 3 x - 2$ is $(x + a) (2 x - b) (3 x + c)^{2}$ . Then $a - b - c =$ _____

$3$

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$6$

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$0$

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$9$

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Solution

The correct option is C
$0$

First expression:
$9 x^{2} + 6 x + 1 = (3 x + 1)^{2}$

Second expression:
$3 x^{2} + 7 x + 2 = 3 x^{2} + 6 x + x + 2$
$= 3 x (x + 2) + 1 (x + 2)$
$= (3 x + 1) (x + 2)$

Third expression:
$2 x^{2} + 3 x - 2 = 2 x^{2} + 4 x - x - 2$
$= 2 x (x + 2) - 1 (x + 2)$
$= (2 x - 1) (x + 2)$

$∴$ Required L.C.M. is $(x + 2) (2 x - 1) (3 x + 1)^{2}$

$⟹ a = 2, b = 1, c = 1$

So, $a - b - c = 0$

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