A B 3

$\\Solution:\\ \\Solution:\\\\ Given:\ (a + b)= 6\ \ \ and\ \ \ (ab)= 8\\\\ \text{We have the identity:}\\\\ (a+b)^3=a^3+b^3+3ab(a+b)\\\\ \Rightarrow (a^3+b^3)=(a+b)^3-3ab(a+b)\\\\ \Rightarrow (a^3+b^3)=6^3-3\times8\times6\\\\ \Rightarrow (a^3+b^3)=216-144\\\\ \therefore (a^3+b^3)=72\\$

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