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Question
Find the sum of a2 − ab + bc, 2ab + bc − 2a2 and − 3bc + 3a2 + ab.
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Solution
(a2 − ab + bc) + (2ab + bc − 2a2) + (− 3bc + 3a2 + ab)
= a2 − ab + bc + 2ab + bc − 2a2 − 3bc + 3a2 + ab
= a2 + 3a2 − 2a2 + 2ab + ab − ab + bc + bc − 3bc
= 4a2 − 2a2 + 3ab − ab + 2bc − 3bc
= 2a2 + 2ab − bc
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