Question
Prove that $(a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b )(b + c)(c + a).$
✓
Answer
$(a + b + c)^3$
$= [a + (b + c)]^{3} $
$= a^3 + 3a^2(b + c) + 3a(b + c)^2 + (b + c)^3$
$= a^3 + 3a^2b + 3a^2c + 3a(b^2 + 2bc + c^2) + (b^3 + 3b^2c + 3bc^2 + c^3)$
$= a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^{3}$
$= a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2c + 3c^2a + 3c^2b + 6abc$
$= a^3 + b^3 + c^3 + 3a^2(b + c) + a^3 + b^3 + c^3 + 3a^2(b + c)$
Hence, above result can be put in the form ($a + b + c)^{3}$
$= (a + b + c)^3 + 3(a + b)(b + c)(c + a)$
$\therefore (a + b + c)^3 - a^3 - b^3 - c^3$
$= 3(a + b)(b + c)(c + a)$
$= [a + (b + c)]^{3} $
$= a^3 + 3a^2(b + c) + 3a(b + c)^2 + (b + c)^3$
$= a^3 + 3a^2b + 3a^2c + 3a(b^2 + 2bc + c^2) + (b^3 + 3b^2c + 3bc^2 + c^3)$
$= a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^{3}$
$= a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2c + 3c^2a + 3c^2b + 6abc$
$= a^3 + b^3 + c^3 + 3a^2(b + c) + a^3 + b^3 + c^3 + 3a^2(b + c)$
Hence, above result can be put in the form ($a + b + c)^{3}$
$= (a + b + c)^3 + 3(a + b)(b + c)(c + a)$
$\therefore (a + b + c)^3 - a^3 - b^3 - c^3$
$= 3(a + b)(b + c)(c + a)$
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