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

(N/A) We know the expansion of $(a+b+c)^3$ is given by:
$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$
Alternatively,we can expand it step-by-step:
$(a+b+c)^3 = [(a+b)+c]^3$
$= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Now,subtract $(a^3 + b^3 + c^3)$ from both sides:
$(a+b+c)^3 - a^3 - b^3 - c^3 = 3a^2b + 3ab^2 + 3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2$
Factor out $3$ from the right side:
$= 3[a^2b + ab^2 + a^2c + 2abc + b^2c + ac^2 + bc^2]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + 2abc]$
$= 3[ab(a+b) + ac(a+c) + bc(b+c) + abc + abc]$
$= 3[ab(a+b) + abc + ac(a+c) + ac^2 + bc(b+c) + abc]$
$= 3[ab(a+b+c) + ac(a+c+b) + bc(b+c+a)]$
$= 3(a+b+c)(ab + ac + bc)$
Wait,the standard identity is $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$.
Let's verify: $3(a+b)(b+c)(c+a) = 3(ab + ac + b^2 + bc)(c+a)$
$= 3(abc + a^2b + ac^2 + a^2c + b^2c + ab^2 + bc^2 + abc)$
$= 3(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc)$.
This matches the expansion derived above.
Hence,$(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)$ is proved.