jmc

Let $$p(x)=\frac{(x+a{)}^3}{(a-b)(a-c)}+\frac{(x+b{)}^3}{(b-a)(b-c)}+\frac{(x+c{)}^3}{(c-a)(c-b)}.$$Then $$\begin{array}{rl}p(-a)& =\frac{(-a+a{)}^3}{(a-b)(a-c)}+\frac{(-a+b{)}^3}{(b-a)(b-c)}+\frac{(-a+c{)}^3}{(c-a)(c-b)}\\ & =\frac{(b-a{)}^3}{(b-a)(b-c)}+\frac{(c-a{)}^3}{(c-a)(c-b)}\\ & =\frac{(b-a{)}^2}{b-c}+\frac{(c-a{)}^2}{c-b}\\ & =\frac{(b-a{)}^2-(c-a{)}^2}{b-c}\\ & =\frac{[(b-a)+(c-a)][(b-a)-(c-a)]}{b-c}\\ & =\frac{(b+c-2a)(b-c)}{b-c}\\ & =b+c-2a\\ & =(a+b+c)+3(-a)\end{array}$$Similarly, $$\begin{array}{rl}p(-b)& =a+c-2b=(a+b+c)+3(-b),\\ p(-c)& =a+b-2c=(a+b+c)+3(-c).\end{array}$$Since $p(x)=a+b+c+3x$ for three distinct values of $x,$ by the Identity Theorem, $p(x)=\boxed{a+b+c+3x}$ for all $x.$