Prove $3a^4-4a^3b+b^4\ge0$ . $(a,b\in\mathbb{R})$ $$3a^4-4a^3b+b^4\ge0\ \ (a,b\in\mathbb{R})$$
We must factorize $3a^4-4a^3b+b^4\ge0$ and get an expression with an even power like $(x+y)^2$ and say an expression with an even power can not have a negative value in $\mathbb{R}$. 
But I don't know how to factorize it since it is not in the shape of any standard formula.

$$3a^4-4a^3b+b^4$$
$$=3a^4-3a^3b-a^3b+b^4$$
$$=3a^3(a-b)-b(a^3-b^3)$$
$$=(a-b)\Big(3a^3-b(a^2+b^2+ab)\Big)$$
$$=(a-b)(3a^3-a^2b-b^3-ab^2)$$
The cubic is zero if $a=b$, so try taking $(a-b)$ out again:
$$(a-b)(3a^3-3a^2b+2a^2b-2b^3+b^3-ab^2)$$
$$=(a-b)\Big(3a^2(a-b)+2b(a^2-b^2)-b^2(a-b)\Big)$$
$$=(a-b)^2(3a^2+b^2+2ab)$$
$$=(a-b)^2\Big((a+b)^2+2a^2\Big)$$
