An Algebraic Challenge -- Part 3
If you haven't tried An Algebraic Challenge -- Part 2 , you may try it.
Before looking at the question, please read the instruction carefully.
INSTRUCTION: The below given question is just for self analysis of how much time you take to solve it. If you can solve the question WITHIN 18 SECONDS after reading the QUESTION, then, you are exceptional (according to me).
QUESTION: If a² + 2b = 7, b² + 4c = -7 and c² + 6a = -14, what is the value of (-abc)?
The answer is 6.
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1 solution
a² + 2b = 7, b² + 4c = -7 and c² + 6a = -14
Adding all the three equations, we get
a² + b² + c² +2b + 4c + 6a = 7 - 7 -14
a² + b² + c² +2b + 4c + 6a = -14
a² + b² + c² +2b + 4c + 6a + 14 = 0
(a² + 6a +9) + (b² + 2b + 1) + (c² + 4c + 4) = 0
(a + 3)² + (b + 1)² + (c + 2)² = 0
But, this is only possible if a + 3 = 0, a = -3
b + 1 = 0, b = -1
c + 2 = 0, c = -2
(-abc) = -(-3)(-1)(-2) = 6