Question

if x+y+z=0 then the value of (x^2y^2+x^2z^2+z^2 y^2)/(x^4+y^4+z^4)

Answer 1

The answer is ½.

The easy way to solve this is to give values to x and y. Let x=1 and y=2, then z=-3 because z=-(x+y).

x²y²+x²z²+y²z²=4+9+36=49; x⁴+y⁴+z⁴=1+16+81=98; 49/98=½.

Now let x=2, y=3, so z=-5:

x²y²+x²z²+y²z²=361; x⁴+y⁴+z⁴=1+16+81=722; 361/722=½.

Now to work out the reason:

z=-(x+y), so z²=(x+y)².

x²y²+x²z²+y²z²=x²y²+z²(x²+y²)=x²y²+(x+y)²(x²+y²)=

x²y²+(x+y)²[(x+y)²-2xy]=x²y²-2xy(x+y)²+(x+y)⁴=(xy-(x+y)²)²=(-x²-xy-y²)²=(x²+xy+y²)²=((x²+y²)+xy)²=(x²+y²)²+2xy(x²+y²)+x²y².

x⁴+y⁴+z⁴=x⁴+y⁴+(x+y)⁴=x⁴+y⁴+(x²+2xy+y²)²=x⁴+y⁴+((x²+y²)+2xy)²=

x⁴+y⁴+(x²+y²)²+4xy(x²+y²)+4x²y²=(x²+y²)²-2x²y²+(x²+y²)²+4xy(x²+y²)+4x²y²=2(x²+y²)²+4xy(x²+y²)+2x²y²=

2((x²+y²)²+2xy(x²+y²)+x²y²).

Therefore:

(x²y²+x²z²+y²z²)/(x⁴+y⁴+z⁴)=((x²+y²)²+2xy(x²+y²)+x²y²)/(2((x²+y²)²+2xy(x²+y²)+x²y²))=½.

That is: (x²+xy+y²)²/(2(x²+xy+y²)²)=½.