若x²-ax+3c=12,且a²+b²+c²=ab+bc+ca,则a+b²+c³=
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![若x²-ax+3c=12,且a²+b²+c²=ab+bc+ca,则a+b²+c³=](/uploads/image/z/1835100-36-0.jpg?t=%E8%8B%A5x%26%23178%3B-ax%2B3c%3D12%2C%E4%B8%94a%26%23178%3B%2Bb%26%23178%3B%2Bc%26%23178%3B%3Dab%2Bbc%2Bca%2C%E5%88%99a%2Bb%26%23178%3B%2Bc%26%23179%3B%3D)
若x²-ax+3c=12,且a²+b²+c²=ab+bc+ca,则a+b²+c³=
若x²-ax+3c=12,且a²+b²+c²=ab+bc+ca,则a+b²+c³=
若x²-ax+3c=12,且a²+b²+c²=ab+bc+ca,则a+b²+c³=
a²+b²+c²=ab+bc+ca
∴(a-b)²+(b-c)²+(c-a)²=0
∴a=b=c
a+b²+c^3=a+a²+a^3=a(1+a+a²)
在x²-ax+3c=12中,x²-ax+3a=12,令x=a,则a²-a²+3a=12,∴a=4
∴a+b²+c^3=a(1+a+a²)=4(1+4+16)=84