Answer
424.2k+ views
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 $ .
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 $ .
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)