jmc

Let $a=x-3$ and $b=x-7.$ Then we can write the given equation as $$a^3+b^3=(a+b{)}^3.$$Expanding, we get $a^3+b^3=a^3+3a^{2}b+3a{b}^2+b^3,$ so $3a^{2}b+3a{b}^2=0,$ which factors as $$3ab(a+b)=0.$$Thus, $a=0,$ $b=0,$ or $a+b=0.$ Then $x-3=0,$ $x-7=0,$ or $2x-10=0.$ This gives us the roots $\boxed{3,5,7}.$