Prove that:
$(a + b)^{3} (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)$ .

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Solution

$L . H . S . = {(a + b)}^{3} + {(b + c)}^{3} + {(c + a)}^{3} - 3 (a + b) (b + c) (c + a) = (a + b + b + c + c + c + a) [{(a + b)}^{2} + {(b + c)}^{2} + {(c + a)}^{2} - (a + b) (b + c) - (b + c) (c + a) - (c + a) (a + b)] = (2 a + 2 b + 2 c) (a^{2} + b^{2} + 2 a b + b^{2} + c^{2} + 2 b c + c^{2} + a^{2} + 2 c a - a b - a c - b^{2} - b c - b c - a b - c^{2} - c a - c a - b c - a^{2} - a b) = 2 (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c) = R . H . S . Hence, proved .$

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Similar questions

(a + b)^{3} + (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)

\frac{(a + b)^{3} + (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a)}{3 (a^{3} + b^{3} + c^{3} - 3 a b c)}

{(b + c)}^{3} + {(c + a)}^{3} + {(a + b)}^{3} - 3 (b + c) (c + a) (a + b) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

|\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{matrix}| = a^{3} + b^{3} + c^{3} - 3 a b c

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