Tick (✓) the correct answer:
The sum of (6a + 4b − c + 3), (2b − 3c + 4), (11b − 7a + 2c − 1) and (2c − 5a − 6) is
(a) (4a − 6b + 2)
(b) (−3a + 14b − 3c + 2)
(d) (−6a + 17b)
(d) (−6a + 6b + c −4)

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Solution

(c) (−6a + 17b)

$6 a + 4 b - c + 3 + 2 b - 3 c + 4 - 7 a + 11 b + 2 c - 1 - 5 a + 2 c - 6 \bar{- 6 a + 17 b + 0 c + 0}$

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∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 2 a & 3 a + 2 b & 4 a + 3 b + 2 c 3 a & 6 a + 3 b & 10 a + 6 b + 3 c \end{matrix} ∣ ∣ ∣ ∣

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∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 2 a & 3 a + 2 b & 4 a + 3 b + 2 c 3 a & 6 a + 3 b & 10 a + 6 b + 3 c \end{matrix} ∣ ∣ ∣ ∣ = a^{3}

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