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.

$$\color{red}{3a^4}-4a^3b+b^4=\color{red}{4a^4-a^4}-4a^3b+b^4=4a^3(a-b)+b^4-a^4$$
writing $$b^4-a^4=(b-a)(b+a)(b^2+a^2)$$ and factoring out $(a-b)$ gives
$$4a^3(a-b)+(b-a)(b+a)(b^2+a^2)=(a-b)\color{blue}{(4a^3-(a^3+ab^2+ba^2+b^3\big)})$$
Consider two cases.
1)$a\ge b$. Then the blue color expression is $\ge 0$. Because
$$a^3+ab^2+ba^2+b^3\le a^3+a^3+a^3=4a^3$$
2)$a\le b$. The proof is Similar to the case 1.
