 admin # Determine the sum: (14abc2 + 12a2b + 16b2c) + (−3abc2 + 2b2c). 11abc2 + 12a2b + 18b2c 11a2b2c4 + 12a2b + 18b4c2 41a4b7c6 42a3b4c3 − ab3c3

2 months ago

## Solution 1 Guest #4318
2 months ago
The answer is 11abc^2 + 12a^2b + 18b^2c because:
14abc^2-3abc^2 is 11abc^2
12a^2b + 0 = 12a^2b
16b^2c +2b^2c = 18b^c

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