<s>[INST] Factoring a given quartic form Description
Below is our polynomial:
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4$$
I've gone to almost end of it but,it seems there's a small -or conversely huge- fault in my solving which made  $\mathbf{A}\:and\: \mathbf{B}$ imaginary numbers as I didn't find any two numbers which Could have the sum of $-8xy$ and product of $24x^2y^2$ OR in another sense , I can not find $\mathbf{A}\:and\: \mathbf{B}$.
My work
$$\begin{align}
 & 3x^4-8x^3y+14x^2y^2-8xy^3+3y^4 \\
 & = 3(x^4+y^4)-8xy(x^2+y^2)+14x^2y^2 \\
 & = 3(x^4+y^4+2x^2y^2)-8xy(x^2+y^2)+8x^2y^2\\
 & = 3(x^2+y^2)^2-8xy(x^2+y^2)+8x^2y^2\\
 & = \frac{1}{3}\times[3^2(x^2+y^2)^2-      3\times8xy(x^2+y^2)+24x^2y^2] \\
& =\frac{1}{3}(3(x^2+y^2)+\mathbf{A})(3(x^2+y^2)+\mathbf{B})
\end{align}$$
 [/INST] Assume by symmetry
$$(ax^2+bxy+cy^2)(cx^2+bxy+ay^2)=\\=acx^4+(ab+bc)x^3y+(a^2+b^2+c^2)x^2y^2+(ab+bc)xy^3+(ac)y^4$$
then we need


*

*$ac=3 \implies a=3\, c=1$

*$ab+bc=-8 \implies3b+b=-8\implies b=-2$

*$a^2+b^2+c^2=14 \implies 9+4+1=14$


thus
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4=(3x^2-2xy+y^2)(x^2-2xy+3y^2)$$
</s>