Math Problem Solver

The left-hand side of the equation evaluates to:

\[\begin{align*}& a + b + \left(b + c\right)^{2} - \left(c + a\right)^{3} - 3 \left(a + b\right) \left(b + c\right) \left(c + a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - \left(c + a\right)^{3} - 3 \left(a + b\right) \left(b + c\right) \left(c + a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - \left(a^{3} + c^{3} + 3 a^{2} c + 3 c^{2} a\right) - 3 \left(a + b\right) \left(b + c\right) \left(c + a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - \left(a^{3} + c^{3} + 3 a^{2} c + 3 c^{2} a\right) - \left(3 a + 3 b\right) \left(b + c\right) \left(c + a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - \left(a^{3} + c^{3} + 3 a^{2} c + 3 c^{2} a\right) - \left(3 a b + 3 a c + 3 b^{2} + 3 b c\right) \left(c + a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - \left(a^{3} + c^{3} + 3 a^{2} c + 3 c^{2} a\right) - \left(6 a b c + 3 c^{2} a + 3 b^{2} c + 3 c^{2} b + 3 a^{2} b + 3 a^{2} c + 3 b^{2} a\right)\\=& a + b + b^{2} + c^{2} + 2 b c - a^{3} - c^{3} - 6 a^{2} c - 6 c^{2} a - 6 a b c - 3 b^{2} c - 3 c^{2} b - 3 a^{2} b - 3 b^{2} a\\=& - a^{3} - 3 a^{2} b - 6 a^{2} c - 3 b^{2} a - 6 a b c - 6 c^{2} a - 3 b^{2} c - 3 c^{2} b - c^{3} + b^{2} + 2 b c + c^{2} + a + b\\\end{align*}\]

The right-hand side of the equation evaluates to:

\[\begin{align*}& 2 \left(a^{3} + b^{3} + c^{3} - 3 a b c\right)\\=& 2 a^{3} + 2 b^{3} + 2 c^{3} - 6 a b c\\=& 2 a^{3} - 6 a b c + 2 b^{3} + 2 c^{3}\\\end{align*}\]

Since $- a^{3} - 3 a^{2} b - 6 a^{2} c - 3 b^{2} a - 6 a b c - 6 c^{2} a - 3 b^{2} c - 3 c^{2} b - c^{3} + b^{2} + 2 b c + c^{2} + a + b \ne 2 a^{3} - 6 a b c + 2 b^{3} + 2 c^{3}$, the equation $a + b + \left(b + c\right)^{2} - \left(c + a\right)^{3} - 3 \left(a + b\right) \left(b + c\right) \left(c + a\right) = 2 \left(a^{3} + b^{3} + c^{3} - 3 a b c\right)$ is not an identity.