ALgebra#2

Note that:

$\begin{aligned} (a+b+c)^3 & = {\color{#3D99F6}a^3+b^3+c^3} + {\color{#D61F06}3(a+b+c)(ab+bc+ca)} \color{#3D99F6} -3abc \\ & = {\color{#3D99F6} 48} + \color{#D61F06}99 \\ & = \boxed{147} \end{aligned}$

${\color{#3D99F6}3(a+b+c)(ab+bc+ca)}=3{\color{#D61F06}(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)}+9{\color{#EC7300}abc}$

$(a+b+c)^3={\color{#20A900}a^3+b^3+c^3}+3{\color{#D61F06}(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)}+6{\color{#EC7300}abc}$

$={\color{#20A900}a^3+b^3+c^3}+3{\color{#D61F06}(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)}+9{\color{#EC7300}abc}-3{\color{#EC7300}abc}$

$=({\color{#20A900}a^3+b^3+c^3}-3{\color{#EC7300}abc})+{\color{#3D99F6}3(a+b+c)(ab+bc+ca)}=48+99=147$

Note that $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 -ab - bc -ac).$

So $(a+b+c)(a^2 + b^2 + c^2 -ab - bc -ac)=48; (a+b+c)(ab+bc+ac)=33$

$(ab+bc+ac)=\dfrac{33}{a+b+c}.$

$\implies (a+b+c)(a^2 + b^2 + c^2) -(a+b+c)(ab+bc+ac) = 48$ $\implies (a+b+c)(a^2 + b^2 + c^2) - 33=48$ $\implies (a+b+c)(a^2 + b^2 + c^2 + 2(ab+bc+ac)) = 81 + 66$ $\implies (a+b+c)^3=\boxed{147}.$