Factorise : $1 - (2 x + 3 y) - 6 (2 x + 3 y)^{2}$

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Solution

$L e t (2 x + 3 y) = a ∴ (2 x + 3 y)^{2} = a^{2} ∴ 1 - (2 x + 3 y) - 6 (2 x + 3 y)^{2} = 1 - a - 6 a^{2} = 1 - 3 a + 2 a - 6 a^{2} = 1 (1 - 3 a) + 2 a (1 - 3 a) = (1 - 3 a) (1 + 2 a) = [1 - 3 (2 x + 3 y)] [1 + 2 (2 x + 3 y)] Substituting the value of a = (1 - 6 x - 9 y) (1 + 4 x + 6 y)$

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