Question

Which of the the following Expressions,are cyclic polynomials.

• A. Both $1$ and $2$
• B. Only $2$
• C. Only $4$
• D. Both $1$, $2$ and $4$

Answer 1

Answer: (D)

Both $1$, $2$ and $4$

A polynomial is said to be cyclic if $f(x,y,z)=f(y,z,x)$
$\left(1\right)x+y+z$
We have $f(x,y,z)=x+y+z$
$\Rightarrow f(y,z,x)=y+z+x=x+y+z=f(x,y,z)$
Therefore $\left(1\right)$ is cyclic.
$\left(2\right)(x-y{)}^2+(y-z{)}^2+(z-x{)}^2$
We have $f(x,y,z)=(x-y{)}^2+(y-z{)}^2+(z-x{)}^2$
$\Rightarrow f(y,z,x)=(y-z{)}^2+(z-x{)}^2+(x-y{)}^2=f(x,y,z)=(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=f(x,y,z)$
Therefore $\left(2\right)$ is cyclic.
$\left(3\right)x(y+z)-y(z+x)+z(x+y)$
We have $f(x,y,z)=x(y+z)-y(z+x)+z(x+y)$
$\Rightarrow f(y,z,x)=y(z+x)-z(x+y)+x(y+z)\ne f(x,y,z)$
Therefore $\left(3\right)$ is not cyclic.
$\left(4\right)x^2+y^2+z^2$
We have $f(x,y,z)=x^2+y^2+z^2$
$\Rightarrow f(y,z,x)=y^2+z^2+x^2=x^2+y^2+z^2=f(x,y,z)$
Therefore $\left(4\right)$ is cyclic.
Hence $\left(1\right),\left(2\right),\left(4\right)$ are cyclic.