Factorise:
$9 a^{2} - 4 b^{2} - 6 a + 1$

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Solution

Given: $9 a^{2} - 4 b^{2} - 6 a + 1$
$= 9 a^{2} - 6 a + 1 - 4 b^{2}$
$= {(3 a)^{2} - 2 (3 a) + 1^{2}} - (2 b)^{2}$
$= (3 a - 1)^{2} - (2 b)^{2}$ [ $∵ (a^{2} - 2 a b + b^{2} = (a - b)^{2}$ ]
$= (3 a + 2 b - 1) (3 a - 2 b - 1)$ [ $∵ (a^{2} - b^{2} = (a + b) (a - b)$ ]

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