Question:
Prove that
$(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$
Solution:
$(a+b+c)^{3}=[(a+b)+c]^{3}=(a+b)^{3}+c^{3}+3(a+b) c(a+b+c)$
$\Rightarrow(a+b+c)^{3}=a^{3}+b^{3}+3 a b(a+b)+c^{3}+3(a+b) c(a+b+c)$
$\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3 a b(a+b)+3(a+b) c(a+b+c)$
$\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)\left[a b+c a+c b+c^{2}\right]$
$\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)[a(b+c)+c(b+c)]$
$\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)(b+c)(a+c)$
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