Question

Show that for all real numbers x, y, z such that x + y + z = 0 and xy + yz + zx = 3. The expression ${{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x$ is constant.

Answer 1

Hint: Assume a cubic equation, which has three roots as ‘x, y, z’. Simplify it by multiplying the brackets of them and using the values of $x+y+z$ and $xy+yz+zx$, as given in the problem. In the obtained equation, substitute x, y, z and you will obtain 3 equations. Now multiply the 3 equations with corresponding values to obtain the expression ${{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x$ . Then solve to get the desired result.

Complete step-by-step answer:
Here, we have
x + y + z = 0…………………….(i)
xy + yz + zx = 3……………….(ii)
Where x ,y ,z are real numbers. We know that roots of the polynomial
$\begin{aligned} & \left( x-{{\alpha }_{1}} \right)\left( x-{{\alpha }_{2}} \right)\left( x-{{\alpha }_{3}} \right)...........\left( x-{{\alpha }_{n}} \right)=0, \\\ & {{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}}.................{{\alpha }_{n}} \\\ \end{aligned}$
So, let us suppose a polynomial which have roots (x, y, z) as
(m – x) (m – y) (m – z) = 0……………………………..(iii)
So, the above equation in variable ‘m’ has roots (x, y, z) by equating (m – x), (m – y) and (m – z) to 0.
Now, we can multiply the brackets of equation (iii) and hence, get
$\begin{aligned} & \left( {{m}^{2}}-mx-my+xy \right)\left( m-z \right)=0 \\\ & {{m}^{3}}-{{m}^{2}}z-m{{x}^{2}}+mxz-{{m}^{2}}y+myz+mxy-xyz=0 \\\ \end{aligned}$
Now, we can rewrite the above equation as
${{m}^{3}}-{{m}^{2}}\left( x+y+z \right)+m\left( xy+yz+zx \right)-xyz=0$
Now, put values of x + y + z and xy + yz + zx as 0 and 3 respectively from the equation (i) and (ii) we get
$\begin{aligned} & {{m}^{3}}-{{m}^{2}}\left( 0 \right)+3m-xyz=0 \\\ & {{m}^{3}}+3m-xy=0..................\left( iv \right) \\\ \end{aligned}$
Now, we know that (x, y, z) are the roots of equation (iv). It means x, y, z will satisfy the equation (iv). Hence, on putting x, y and z to the equation (iv) we get, three equations simultaneously as
$\begin{aligned} & {{x}^{3}}+3x-xyz=0...........\left( v \right) \\\ & {{y}^{3}}+3y-xyz=0............\left( vi \right) \\\ & {{z}^{3}}+3z-xyz=0.............\left( vii \right) \\\ \end{aligned}$
Now, multiply the above equations (v), (vi), (viii) by y, z and x respectively. So, we get all three equations as
$\begin{aligned} & {{x}^{3}}y+3xy-x{{y}^{2}}z=0..............\left( viii \right) \\\ & {{y}^{3}}z+3yz-xy{{z}^{2}}=0..............\left( i\text{x} \right) \\\ & {{z}^{3}}x+3zx-{{x}^{2}}yz=0..............(\text{x}) \\\ \end{aligned}$
Now, add all three equations to determine the value of ${{x}^{3}}y+{{y}^{3}}z+{{x}^{3}}z$ as asked in the problem. So, adding the equations (viii), (ix), (x); we get
$\left( {{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x \right)+3\left( xy+yz+zx \right)-xyz\left( x+y+z \right)=0$
Where sum $x{{y}^{2}}z+xy{{z}^{2}}+{{x}^{2}}yz$ is written ass xyz (x + y + z) by taking xyz common from the expression. So, we can put value of x + y + z and xy + yz + zx from the equations (i) and (ii) we get

& \left( {{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x \right)+3\left( 3 \right)-xyz\left( 0 \right)=0, \\\ & {{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x+9=0, \\\ & {{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x=-9 \\\ \end{aligned}$$ Hence, it is shown that $${{x}^{3}}y+{{y}^{3}}z+{{z}^{3}}x$$ is a constant value if x + y + z = 0 and xy + yz + zx = 3. Note: Another approach for solving the question would be that we can find the value of any two variables in terms of other variables. It is a very complex approach and will take time, but the answer will remain the same. One may assume xyz as any other variable (For example: $\lambda $ ) for not getting confused with equations $${{x}^{3}}+3x-xyz=0,{{y}^{3}}+3y-xyz+{{x}^{3}}y,{{z}^{3}}+3z-xyz=0.$$ As, one may cancel (x, y, z) from the equations and get the value as 0. So, don’t get confused here and put xyz as any other variable, example; $xyz=\lambda $. One cannot find the exact value of x, y, z as we have only two equations but 3 variables. So, don’t go for calculating the exact values of x. y ,z.