I'm having trouble with 3 variable systems of equations, can you help
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I'm having trouble with 3 variable systems of equations, can you help

[From: ] [author: ] [Date: 12-09-11] [Hit: ]
and c. Please show steps as to how you got the answers.abc = -144-Note that the equations are completely symmetric in a, b, and c. That means that if a=α,......
The goal is to solve for variables a, b, and c. Please show steps as to how you got the answers.
a+b+c = 3
ab+bc+ac = 0
abc = -144

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Note that the equations are completely symmetric in a, b, and c. That means that if a=α, b=β, and c=γ is a solution, then every permutation of (α,β,γ) is also a solution (e.g., a=β, b=α, and c=γ is also a solution).

If you expand (x-a)(x-b)(x-c) = 0, you get

x³ - (a + b + c)x² + (ab + bc + ac)x - abc = 0

Using your equations, this leads to

x³ - 3x² + 144 = 0

The solutions to this equation are the solutions for a, b, and c (because (x-a)(x-b)(x-c) = 0 implies x=a or x=b or x=c). I submitted the equation to WolframAlpha, and it has one real solution and a complex-conjugate pair of solutions. You can see them here: http://www.wolframalpha.com/input/?i=sol…
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