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.
 A: since AM-GM inequality $$3a^4+b^4=a^4+a^4+a^4+b^4\ge 4(a^{12}b^4)^{1/4}=4|a|^3|b|\ge 4a^3b$$
A: $$3a^4-4a^3b+b^4$$
$$=3a^4-3a^3b-a^3b+b^4$$
$$=3a^3(a-b)-b(a^3-b^3)$$
$$=(a-b)\Big(3a^3-b(a^2+b^2+ab)\Big)$$
$$=(a-b)(3a^3-a^2b-b^3-ab^2)$$
The cubic is zero if $a=b$, so try taking $(a-b)$ out again:
$$(a-b)(3a^3-3a^2b+2a^2b-2b^3+b^3-ab^2)$$
$$=(a-b)\Big(3a^2(a-b)+2b(a^2-b^2)-b^2(a-b)\Big)$$
$$=(a-b)^2(3a^2+b^2+2ab)$$
$$=(a-b)^2\Big((a+b)^2+2a^2\Big)$$
A: 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.
A: $$\color{red}{3a^4}-4a^3b+b^4=\color{red}{4a^4-a^4}-4a^3b+b^4=4a^3(a-b)+b^4-a^4$$
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.
