Expand: ${(3 a - 4 b - 3 c)}^{2}$

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Solution

We have ${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$

Take $x = 3 a, y = - 4 b$ and $z = - 3 c$

${(3 a - 4 b - 3 c)}^{2} = {(3 a)}^{2} + {(- 4 b)}^{2} + {(- 3 c)}^{2} + 2 (3 a \times - 4 b + (- 4 b) \times - 3 c + (- 3 c) \times 3 a)$

${(3 a - 4 b - 3 c)}^{2} = 9 a^{2} + 16 b^{2} + 9 c^{2} + 2 (- 12 a b + 12 b c - 9 a c)$

$∴ {(3 a - 4 b - 3 c)}^{2} = 9 a^{2} + 16 b^{2} + 9 c^{2} - 24 a b + 24 b c - 18 a c$

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