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

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Solution

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

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

${x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2}) x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})}$

$= (a + b + c - a) [a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a + a^{2} + a b + c a + a^{2}]$

$- (b + c) (b^{2} - b c + c^{2})$

${{(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x}$

$= (b + c) [3 a^{2} + b^{2} + c^{2} + 3 a b + 2 b c + 3 c a]$

$- (b + c) (b^{2} - b c + c^{2})$

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

$= (b + c) [3 a^{2} + 3 a b + 3 b c + 3 c a]$

$= 3 (b + c) [a (a + b) + c (a + b)]$

$= 3 (a + b) (b + c) (c + a)$ = RHS

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Similar questions

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

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

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

(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)}

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

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