Question 338934
Use a substitution.
Let {{{D=b+c}}}.
{{{(a+b+c)(a+b+c)=(a+D)(a+D)}}}
{{{(a+b+c)(a+b+c)=a^2+aD+Da+D^2}}}
{{{(a+b+c)(a+b+c)=a^2+2aD+D^2}}}
Now substitute back,
{{{(a+b+c)(a+b+c)=a^2+a(b+c)+(b+c)a+(b+c)^2}}}
{{{(a+b+c)(a+b+c)=a^2+ab+ac+ab+ac+b^2+2bc+c^2}}}
{{{(a+b+c)(a+b+c)=a^2+b^2+c^2+2ab+2ac+2bc}}}
{{{(a+b+c)(a+b+c)=a^2+b^2+c^2+2(ab+bc+ac)}}}