Question

If $\dfrac{{{x}^{2}}-yz}{a}=\dfrac{{{y}^{2}}-zx}{b}=\dfrac{{{z}^{2}}-xy}{c}$, show that $\left( x+y+z \right)\left( a+b+c \right)=ax+by+cz.$ \[\]

Answer 1

Hint: For solving this question we have to first of all take the given equation we will equate it with any non-zero constant $k$ and then after rearranging we will we the formula otherwise known as ${{8}^{\text{th}}}$identity of algebraic factorization that is ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ .After using this formula we will be prove the above that $\left( x+y+z \right)\left( a+b+c \right)=ax+by+cz.$\[\]

Complete step-by-step answer:
We are given the following continued proportion of polynomials
\[\dfrac{{{x}^{2}}-yz}{a}=\dfrac{{{y}^{2}}-zx}{b}=\dfrac{{{z}^{2}}-xy}{c}\]
Let us consider for any none zero constant $k$ we have
\[\dfrac{\left( {{x}^{2}}-yz \right)}{a}=\dfrac{\left( {{y}^{2}}-zx \right)}{b}=\dfrac{\left( {{z}^{2}}-xy \right)}{c}=k\]
We equate each term in the above proportion with $k$ to have
\[\dfrac{\left( {{x}^{2}}-yz \right)}{a}=k\Rightarrow {{x}^{2}}-yz=ak,\dfrac{\left( {{y}^{2}}-zx \right)}{b}=k\Rightarrow {{y}^{2}}-zx=bk,\dfrac{\left( {{z}^{2}}-xy \right)}{c}=k\Rightarrow {{z}^{2}}-xy=ck....\left( 1 \right)\]
We take the first term and multiply $x$in the numerator and denominator and then express the numerator ${{x}^{2}}-yz$ in terms of $k$ as;
\[\dfrac{x\left( {{x}^{2}}-yz \right)}{ax}=k\Rightarrow {{x}^{2}}-yz=akx...\left( 2 \right)\]
We take the second term and multiply $y$in the numerator and denominator and then express the numerator ${{y}^{2}}-zx$ in terms of $k$ as;
\[\dfrac{y\left( {{y}^{2}}-zx \right)}{by}=k\Rightarrow y\left( {{y}^{2}}-zx \right)=bky.....\left( 3 \right)\]
We take the third term and multiply $z$in the numerator and denominator and then express the numerator ${{z}^{2}}-xy$ in terms of $k$ as;
 \[\dfrac{z\left( {{z}^{2}}-xy \right)}{cz}=k\Rightarrow z\left( {{z}^{2}}-xy \right)=ckz......\left( 4 \right)\]
We add the terms add respective of equations (2) ,(3) and (4) and to have;
\[x\left( {{x}^{2}}-yz \right)+y\left( {{y}^{2}}-zx \right)+z\left( {{z}^{2}}-xy \right)=kax+kby+kcz\]
We open the brackets in left hand side and take $k$ common in the right hand side in the above step to have
\[\begin{align}
  & {{x}^{3}}-xyz+{{y}^{3}}-xyz+{{z}^{3}}-xyz=k\left( ax+by+cz \right) \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=k\left( ax+by+cz \right) \\
\end{align}\]
We use the algebraic identity ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ in the left hand side of the above step to have;
\[\begin{align}
  & \Rightarrow \left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)=k\left( ax+by+cz \right) \\
 & \Rightarrow \left( x+y+z \right)\left( {{x}^{2}}-xy+{{y}^{2}}-zx+{{z}^{2}}-xy \right)=k\left( ax+by+cz \right) \\
\end{align}\]
We put the expressions in terms of $k$ obtained from equation (1) in the above step to have;
\[\begin{align}
  & \Rightarrow \left( x+y+z \right)\left( ka+kb+kc \right)=k\left( ax+by+cz \right) \\
 & \Rightarrow \left( x+y+z \right)k\left( a+b+c \right)=k\left( ax+by+cz \right) \\
\end{align}\]
Since $k$ is non-zero we can divide both sides of the above step by $k$ to have;
\[\Rightarrow \left( x+y+z \right)\left( a+b+c \right)=ax+by+cz\]
Hence the given statement is proved. \[\]

Note: We know that the ratio among three terms $a:b:c$ is called a continued portion and a proportion of the type $a:b::c:d$ is called continued proportion. We also know that none of the terms in proportion are zero or undefined hence $a,b,c$ cannot be zero. If $x=y+z=0$ then we can find sum of cubes as ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ and also if ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ then $x+y+z=0$.We note that the polynomial of the type $P\left( x,y,z \right)=P\left( y,z,x \right)=P\left( z,x,y \right)$ is called a cyclic polynomial . Here ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx$ is a cyclic polynomial.