Find the square of: $2 a - b - 3 c$

4 a^{2} + b^{2} + 9 c^{2} - 4 a b + 6 b c - c a

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4 a^{2} + b^{2} + 9 c^{2} - 4 a b + b c - 12 c a

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4 a^{2} + b^{2} + 9 c^{2} - 4 a b + 6 b c - 12 c a

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4 a^{2} + b^{2} + c^{2} - 4 a b + 6 b c - 12 c a

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Solution

The correct option is C $4 a^{2} + b^{2} + 9 c^{2} - 4 a b + 6 b c - 12 c a$
$2 a - b - 3 c$
Squaring, we get
$= (2 a - b - 3 c)^{2}$
$= (2 a)^{2} + (- b)^{2} + (- 3 c)^{2} + 2 (2 a) (- b) + 2 (- b) (- 3 c) + 2 (- 3 c) (2 a)$
$= 4 a^{2} + b^{2} + 9 c^{2} - 4 a b + 6 b c - 12 c a$

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Similar questions

4 a^{2} + b^{2} + 9 c^{2} - 4 a b - 6 b c + 12 c a

4 a^{2} - 4 a b + b^{2} - 9 c^{2} + 12 c d - 4 d^{2}

4 a^{2} + 4 a b + b^{2} =

Factorise: $4 a^{2} + 4 a b + b^{2}$

(2 a + b)^{2} = 4 a^{2} + b^{2} + 4 a b

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