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Given: $ {{z}_{1}}+{{z}_{2}}+{{z}_{3}}=A;{{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}}=B;{{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w=C $ , where ‘w’ is the cube root of unity. Prove: $ {{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\} $ .

Answer Verified Verified
Hint: At first take the absolute value of the square of A, B, C using the formula $ {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca. $ Now add them up and use the fact $ {{w}^{3}}=1,\left| w \right|=1,1+w+{{w}^{2}}=0 $ to get the desired result.

Complete step-by-step answer:
In the question given that \[A={{z}_{1}}+{{z}_{2}}+{{z}_{3}},B={{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}}\] and $ C={{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w $ also ‘w’ is the cube root of 1, so we can write $ {{w}^{3}}=1 $ .
Now we will use formula,
 $ {{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \right)..........(i) $
At first, we will prove it by,
\[{{\left( a+b+c \right)}^{2}}=\left( a+b+c \right)\left( a+b+c \right)\]
Now multiplying it, we will get,
\[a\left( a+b+c \right)+b\left( a+b+c \right)+c\left( a+b+c \right)\]
\[\Rightarrow {{a}^{2}}+ab+ac+{{b}^{2}}+ab+bc+ac+bc+{{c}^{2}}\]
Now, adding and rearranging, we will get,
\[\Rightarrow {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\]
So, now using it we will find the value of $ {{\left| A \right|}^{2}} $ ,
 $ {{\left| A \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)}^{2}} $
Now using the formula from equation (i), we get
 $ {{\left| A \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}+2{{z}_{1}}{{z}_{2}}+2{{z}_{2}}{{z}_{3}}+2{{z}_{3}}{{z}_{1}}.........(ii) $
Now using it we will find the value of $ {{\left| B \right|}^{2}} $ ,
\[{{\left| B \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}} \right)}^{2}}\]
Now using the formula from equation (i), we get
 $ {{\left| B \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{4}}+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}{{w}^{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}} $
Now using the fact $ {{w}^{3}}=1 $ and then rearranging the terms, the above equation can be written as,
 $ {{\left| B \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}}...........(iii) $
Now using it we will find the value of $ {{\left| C \right|}^{2}} $ ,
\[{{\left| C \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w \right)}^{2}}\]
Now using the formula from equation (i), we get
 $ {{\left| C \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{4}}+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{2}}+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{2}}{{z}_{3}}{{w}^{3}}+2{{z}_{1}}{{z}_{3}}w $
Now using the fact $ {{w}^{3}}=1 $ and then rearranging the terms, the above equation can be written as,
 $ {{\left| C \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}\left| w \right|+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{1}}{{z}_{3}}w+2{{z}_{2}}{{z}_{3}}.......(iv) $
Now we will be adding $ {{\left| A \right|}^{2}},{{\left| B \right|}^{2}},{{\left| C \right|}^{2}} $ we get,
 $ {{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}} $
Substituting values from equation (ii), (iii) and (iv), we get
 $ \begin{align}
  & ={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}+2{{z}_{1}}{{z}_{2}}+2{{z}_{2}}{{z}_{3}}+2{{z}_{3}}{{z}_{1}}+{{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}} \\
 & +{{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}\left| w \right|+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{2}}+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{1}}{{z}_{3}}w+2{{z}_{2}}{{z}_{3}} \\
\end{align} $
Now by rearranging and using the fact which is $ \left| w \right|=1 $ we get,
\[{{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\}+2{{z}_{1}}{{z}_{2}}\left( 1+w+{{w}^{2}} \right)+2{{z}_{1}}{{z}_{3}}\left( 1+w+{{w}^{2}} \right)+2{{z}_{3}}{{z}_{1}}\left( 1+w+{{w}^{2}} \right)\]So, now using fact $ \left( 1+w+{{w}^{2}} \right)=0 $ we get that,
 $ {{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\} $
Hence proved

Note: Students while breaking the square should be cautious as there are high chances of making mistakes of leaving any term. While adding they should also keep the fact such as $ {{w}^{3}}=1,{{w}^{2}}+w+1=0 $ . Students often make mistakes when considering the fact $ \left( 1+w+{{w}^{2}} \right)=0 $ .