Math Problem Statement

已知：a不等于b，a^2+b=b^2+c=c^2+a

求证：2(a+b+c)=a^2+b^2+c^2+3(bc+ca+ab)

Solution

我们先从已知条件开始分析。给出的等式是：

a2+b=b2+c=c2+aa^2 + b = b^2 + c = c^2 + a

为方便处理，我们可以将这个式子分为三个部分来分析。

第一步：将已知的等式分解

我们分别写出三个等式：

a2+b=b2+c=c2+aa^2 + b = b^2 + c = c^2 + a

等式一：a2+b=b2+ca^2 + b = b^2 + c

等式二：b2+c=c2+ab^2 + c = c^2 + a

第二步：化简等式

我们先从等式一入手：

a2+b=b2+ca^2 + b = b^2 + c

将右边的 b2+cb^2 + c 移到左边，得：

a2−b2=c−ba^2 - b^2 = c - b

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Math Problem Analysis

Mathematical Concepts

Algebra
Simultaneous Equations
Polynomial Equations

Formulas

a^2 + b = b^2 + c = c^2 + a
2(a+b+c) = a^2 + b^2 + c^2 + 3(bc+ca+ab)

Theorems

Symmetry in Equations
Polynomial Identity

Suitable Grade Level

Grades 10-12

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