Factor (x^2 + xy + y^2)^2 - 4xy(x^2 + y^2) completely.

Step-by-step solution

1. 1

   Expand the square carefully. Using $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$ with $a=x^{2}$, $b=xy$, $c=y^{2}$:

   $\left(x^{2}+xy+y^{2}\right)^{2}=x^{4}+x^{2}y^{2}+y^{4}+2x^{3}y+2x^{2}y^{2}+2xy^{3}=x^{4}+2x^{3}y+3x^{2}y^{2}+2xy^{3}+y^{4}.$

   Note $2ac=2x^{2}y^{2}$ combines with $b^{2}=x^{2}y^{2}$ to give the coefficient $3$.

2. 2

   Subtract the second piece. Since $4xy\left(x^{2}+y^{2}\right)=4x^{3}y+4xy^{3}$,

   $x^{4}+2x^{3}y+3x^{2}y^{2}+2xy^{3}+y^{4}-4x^{3}y-4xy^{3}=x^{4}-2x^{3}y+3x^{2}y^{2}-2xy^{3}+y^{4}.$

   Only the two odd-degree cross terms change sign; the result is still symmetric in $x$ and $y$ and palindromic in its coefficients $1,-2,3,-2,1$.

3. 3

   Exploit the palindromic structure. A symmetric quartic with equal outer coefficients is a natural candidate for

   $\left(x^{2}+axy+y^{2}\right)\left(x^{2}+bxy+y^{2}\right)=x^{4}+(a+b)x^{3}y+(ab+2)x^{2}y^{2}+(a+b)xy^{3}+y^{4}.$

   The $x^{3}y$ and $xy^{3}$ coefficients coming out equal is exactly what the target has, so this ansatz is the right shape.

4. 4

   Match coefficients and solve. Comparing with $1,-2,3,-2,1$:

   $a+b=-2,\qquad ab+2=3\ \Rightarrow\ ab=1.$

   The pair with sum $-2$ and product $1$ is $a=b=-1$ (the discriminant $4-4=0$ means they coincide), so both factors are the same:

   $x^{4}-2x^{3}y+3x^{2}y^{2}-2xy^{3}+y^{4}=\left(x^{2}-xy+y^{2}\right)^{2}.$

5. 5

   State the identity and check. Therefore

   $\left(x^{2}+xy+y^{2}\right)^{2}-4xy\left(x^{2}+y^{2}\right)=\left(x^{2}-xy+y^{2}\right)^{2}.$

   The factor $x^{2}-xy+y^{2}$ is positive for all real $(x,y)\ne(0,0)$, so the whole expression is a perfect square and never negative. Numerical check at $(x,y)=(2.4,1.1)$: both sides give $18.7489$ ✓, and three further random pairs agree to nine digits.