2021/027) Show that for all real numbers $x,\,y,\,z$ such that $x+y+z=0$ and $xy+yz+zx=-3$, the expression $x^3y+y^3z+z^3x$ is a constant.

We have (given)

$x+y+z=0 \cdots(1)$

$xy+yz+zx=-3\cdots(2)$

From (1)

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy +yz + zx) = 0$

or $x^2 + y^2 + z^2 + 2(-3) = 0$

or $x^2 + y^2 + z^2 = 6\cdots(3)$

Further from (1)

$x+y = -z\cdots(4)$

$y+z = -x\cdots(5)$

$z+x = -y\cdots(6)$

Now let us prove that

$x^3y+ y^3 z + z^3 x = xy^3 + yz^3 + zx^3\cdots(7)$

to prove the same

$x^3y+ y^3 z + z^3 x - (xy^3 + yz^3 + zx^3)$

$= (x^3y - xy^3) + (y^3z - yz^3) + (z^3 x - zx^3)$

$= xy(x^2 - y^2) + yz(y^2 - z^2) + zx(z^2 - x^2)$

$=xy(x+y)(x-y) + yz(y+z)( y-z) + zx(z+x)(z-x)$

$=xy(-z)(x-y) + yz(-x) (y-z) + xz(-y) (z-x)$ using (4), (5), (6)

$= - xyz(x-y) - xyz(y-z) - xyz(z-x)$

= 0

so (7) is true

Now $(x^2 + y^2 + z^2)(xy + yz + zx) = 6 * (-3)$ putting values  from above

Or $x^3y + x^2yz + zx^3 + xy^3 + y^3 z + y^2zx + z^2yx + yz^3 + z^3 x = 18$

or $(x^3y + y^3 z + z^3x) + (xy^3 + yz^3 + zx^3 ) + (x^2yz + xy^2z + xyz^2) = - 18$

or  $(x^3y + y^3 z + z^3x) + (x^3y + y^3z + z^3x ) + (x^2yz + xy^2z + xyz^2) = - 18$ (from (7)

or  $2(x^3y + y^3 z + z^3x) + xyz(x+y+z) = - 18$

or $2(x^3y + y^3 z + z^3x) + xyz. 0 = - 18$ from (1)

or $2(x^3y + y^3 z + z^3x)=  - 18$ from (1)

or $(x^3y + y^3 z + z^3x)  = - 9$

Which is a constant

Hence proved