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$\\Solution\\\\ Considering\ LHS\\\\ =(a+b)^3+c^3-a^3-b^3-c^3\\\\ =(a+b)^3+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3\ \ \ \ \ \ [(a+b)^3=a^3+b^3+3ab(a+b)]\\\\ =a^3+b^3+3ab(a+b)+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3\\\\ =a^3+b^3+c^3+3ab(a+b)+3c(a+b)(a+b+c)-a^3-b^3-c^3\\\\ =3ab(a+b)+3c(a+b)(a+b+c)\\\\ = 3(a+b)[ab+c(a+b+c)]\\\\ =3(a+b)[ab+ac+bc+c^2] \\\\ =3(a+b)[a(b+c)+c(b+c)]\\\\ =3(a+b)(b+c)(c+a)=RHS\\\\ Hence\ proced$

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$\\(a+b+c)3 - a^3 - b^3 - c^3 \\= a^3 + b^3 + c^3 + 3ab(a+b) + 3bc(b+c) + 3ac(a+c) + 6abc -a^3 -b^3 -c^3 \\= 3ab(a+b) + 3bc(b+c) + 3ac(a+c) + 6abc \\= 3( ab(a+b) + bc(b+c) + ac(a+c) + 2abc) \\= 3( ab(a+b) + b^2c + abc + bc^2 + ac^2 + a^2c + abc) \\=3( ab(a+b) +bc(a+b) +c2(a+b) + ac(a+b)) \\=3(a+b)(c^2 + bc + ab + ac) \\=3(a+b)( c(b+c) + a(b+c)) \\=3(a+b)(b+c)(c+a)$

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