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a + b = 2 ab = 4 a^2 + b^2 = -4 a^3= ..?
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taking first two equations and combining them: \(a=(2-b)\) substituting in 2º, \((2-b)b=4\) \(-b^2+2b-4=0\) solutions are :\(b_1=1+i\sqrt{3}\) and \(b_2=1-i\sqrt{3}\) now: \(a_1= 2-1-i\sqrt{3}=1-i\sqrt3 \) and \(a_2=2-1+i\sqrt{3}=1+i\sqrt3 \) to get a^3, just solve this: \((1-i\sqrt3)^3=?\)
ahh, fogot to mention. The intuition that there is no real solution, is in a^2 + b^2 = -4 .
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