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.

If $a=b$, the expression is zero, so take the factor $a-b$:
$$3a^4-4a^3b+b^4=(a-b)(3a^3-a^2b-ab^2-b^3)$$
Again, the cubic is zero if $a=b$:
$$=(a-b)^2(3a^2+2ab+b^2)$$
Now try to show the quadratic is always positive or zero.
