Question

If \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\] then
(a) \[a=b=c\]
(b) \[a\ne b\ne c\]
(c) \[a=b\ne c\]
(d) None of the above

Answer 1

Hint: We solve this problem by converting the above equation in such a way that we get sum of positive numbers as zero.
We use the formula of the square of the difference of numbers as
\[{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}\]
Then we use the condition that if the sum of positive numbers is equal to zero then the only possibility is that all the numbers are equal to 0

Complete step by step answer:
We are given that the equation as
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\]
Now, let us try to convert the above equation into a sum of squares of other numbers.
We know that the formula of a square of the difference of numbers as
\[{{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}\]
Here, we can see that we use twice the product of numbers to complete the formula.
Now, let us multiply with 2 on both sides of the given equation then we get
\[\begin{align}
  & \Rightarrow 2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)=2\left( 0 \right) \\
 & \Rightarrow 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0 \\
\end{align}\]
Now, let us rearrange the terms in the above equation such that we get the terms in the required formula then we get
\[\Rightarrow \left( {{a}^{2}}-2ab+{{b}^{2}} \right)+\left( {{b}^{2}}-2bc+{{c}^{2}} \right)+\left( {{c}^{2}}-2ca+{{a}^{2}} \right)=0.........equation(i)\]
Here, we can see that we don’t know about which one of \[a,b,c\] is greater and which is smaller.
But here, we need to use the difference of numbers.
We know that when we do not know about the nature of two numbers then we use modulus for the difference of those two numbers.
Now, by using the above algebra formula and conditions to equation (i) then we get
\[\Rightarrow {{\left( \left| a-b \right| \right)}^{2}}+{{\left( \left| b-c \right| \right)}^{2}}+{{\left( \left| c-a \right| \right)}^{2}}=0\]
Here, we can see that all the three terms are positive because of square.
We know that if the sum of positive numbers is equal to zero then the only possibility is that all the numbers are equal to 0
By using this condition to above equation then we get
\[\Rightarrow \left| a-b \right|=\left| b-c \right|=\left| c-a \right|=0\]
By taking the first term in the above equation then we get
\[\begin{align}
  & \Rightarrow \left| a-b \right|=0 \\
 & \Rightarrow a=b \\
\end{align}\]
Similarly by taking the next term then we get
\[\begin{align}
  & \Rightarrow \left| b-c \right|=0 \\
 & \Rightarrow b=c \\
\end{align}\]
Similarly by taking the next term then we get
\[\begin{align}
  & \Rightarrow \left| c-a \right|=0 \\
 & \Rightarrow c=a \\
\end{align}\]
Therefore we can conclude that the relation as \[a=b=c\]
So, option (a) is the correct answer.


Note:
 We need to note that when we don’t know about the nature of two numbers then we use modulus for a difference of those two numbers.
This is a very important condition.
Let us assume \[x,y\] such that the difference is \[k\]
Here, we don’t know about the nature of \[x,y\] so that there are two possibilities.
(1) \[x-y=k\]
(2) \[y-x=k\]
Here, any possibility may be correct. We know that we cannot take each possibility separately so that we can combine the both possibilities as
\[\begin{align}
  & \Rightarrow \left| x-y \right|=k \\
 & \Rightarrow x-y=\pm k \\
\end{align}\]