Answer
Verified
430.5k+ views
Hint: In this question, we are given an expression as ${{a}^{2}}-ab+{{b}^{2}}$. We have to factorize it. For this, we will use the cube root of unity. Cube root of units means when will take cube root of 1, answer is given by $\omega $. i.e. ${{1}^{\dfrac{1}{3}}}=\omega $ so $1={{\omega }^{3}}$. Using this for 1, we will factorize the given expression. We will also use $1+\omega +{{\omega }^{2}}=0$.
Complete step by step answer:
The expression is ${{a}^{2}}-ab+{{b}^{2}}$.
Since the given expression cannot be factorized as such, we will use cube roots of units.
As we know '$\omega $' is considered as the cube root of unity. Therefore, ${{1}^{\dfrac{1}{3}}}=\omega $.
Taking cube on both sides, we get: ${{\left( {{\left( 1 \right)}^{\dfrac{1}{3}}} \right)}^{3}}={{\omega }^{3}}$.
As we know, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ using it in above equation, we get ${{\left( 1 \right)}^{\dfrac{3}{3}}}={{\omega }^{3}}$.
Simplifying we get: $1={{\omega }^{3}}\cdots \cdots \cdots \left( 1 \right)$.
Also, we know that $1+\omega +{{\omega }^{2}}=0\cdots \cdots \cdots \left( 2 \right)$.
Now, our expression is given as ${{a}^{2}}-ab+{{b}^{2}}$. We need to factorize it using, so we can write it as ${{a}^{2}}+\left( -1 \right)ab+\left( 1 \right){{b}^{2}}\cdots \cdots \cdots \left( * \right)$.
From (2) we know that $1+\omega +{{\omega }^{2}}=0$.
Therefore, $\omega +{{\omega }^{2}}=-1\cdots \cdots \cdots \left( 3 \right)$.
Putting value of -1 and 1 from (3) and (1) respectively in (*) we get:
${{a}^{2}}+\left( \omega +{{\omega }^{2}} \right)ab+{{b}^{2}}{{\omega }^{3}}$.
Simplifying further we get:
${{a}^{2}}+ab\omega +ab{{\omega }^{2}}+{{b}^{2}}{{\omega }^{3}}$.
Taking 'a' common from first two terms and '$b\omega $' common from last two terms, we get:
$a\left( a+b\omega \right)+b{{\omega }^{2}}\left( a+b\omega \right)$.
Taking $\left( a+b\omega \right)$ common from both terms, we get:
$\left( a+b{{\omega }^{2}} \right)\left( a+b\omega \right)$.
Hence, this is the required factored expression.
Note: Formula of $1+\omega +{{\omega }^{2}}=0$ is obtained as following:
As we know, ${{\omega }^{3}}=1$ therefore, ${{\omega }^{3}}-1=0\Rightarrow {{\omega }^{3}}-{{1}^{3}}=0$.
We know that, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ so we get: $\left( \omega -1 \right)\left( {{\omega }^{2}}+\omega +1 \right)=0$.
Now, either $\omega -1=0$ or $\left( {{\omega }^{2}}+\omega +1 \right)$.
But $\omega \ne 1$ so $\omega -1\ne 0$ hence, $\left( {{\omega }^{2}}+\omega +1 \right)$.
Students should take care that we have put $\left( {{\omega }^{2}}+\omega \right)$ in place of -1. Students should note that, for factored expression all the terms should be in product form. Students can simplify expression as $\left( {{a}^{2}}-ab+{{b}^{2}} \right)={{\left( a-b \right)}^{2}}+ab$ and say it as factored form but this is wrong since there is sum of two terms ${{\left( a-b \right)}^{2}}$ but we need products only.
Complete step by step answer:
The expression is ${{a}^{2}}-ab+{{b}^{2}}$.
Since the given expression cannot be factorized as such, we will use cube roots of units.
As we know '$\omega $' is considered as the cube root of unity. Therefore, ${{1}^{\dfrac{1}{3}}}=\omega $.
Taking cube on both sides, we get: ${{\left( {{\left( 1 \right)}^{\dfrac{1}{3}}} \right)}^{3}}={{\omega }^{3}}$.
As we know, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ using it in above equation, we get ${{\left( 1 \right)}^{\dfrac{3}{3}}}={{\omega }^{3}}$.
Simplifying we get: $1={{\omega }^{3}}\cdots \cdots \cdots \left( 1 \right)$.
Also, we know that $1+\omega +{{\omega }^{2}}=0\cdots \cdots \cdots \left( 2 \right)$.
Now, our expression is given as ${{a}^{2}}-ab+{{b}^{2}}$. We need to factorize it using, so we can write it as ${{a}^{2}}+\left( -1 \right)ab+\left( 1 \right){{b}^{2}}\cdots \cdots \cdots \left( * \right)$.
From (2) we know that $1+\omega +{{\omega }^{2}}=0$.
Therefore, $\omega +{{\omega }^{2}}=-1\cdots \cdots \cdots \left( 3 \right)$.
Putting value of -1 and 1 from (3) and (1) respectively in (*) we get:
${{a}^{2}}+\left( \omega +{{\omega }^{2}} \right)ab+{{b}^{2}}{{\omega }^{3}}$.
Simplifying further we get:
${{a}^{2}}+ab\omega +ab{{\omega }^{2}}+{{b}^{2}}{{\omega }^{3}}$.
Taking 'a' common from first two terms and '$b\omega $' common from last two terms, we get:
$a\left( a+b\omega \right)+b{{\omega }^{2}}\left( a+b\omega \right)$.
Taking $\left( a+b\omega \right)$ common from both terms, we get:
$\left( a+b{{\omega }^{2}} \right)\left( a+b\omega \right)$.
Hence, this is the required factored expression.
Note: Formula of $1+\omega +{{\omega }^{2}}=0$ is obtained as following:
As we know, ${{\omega }^{3}}=1$ therefore, ${{\omega }^{3}}-1=0\Rightarrow {{\omega }^{3}}-{{1}^{3}}=0$.
We know that, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ so we get: $\left( \omega -1 \right)\left( {{\omega }^{2}}+\omega +1 \right)=0$.
Now, either $\omega -1=0$ or $\left( {{\omega }^{2}}+\omega +1 \right)$.
But $\omega \ne 1$ so $\omega -1\ne 0$ hence, $\left( {{\omega }^{2}}+\omega +1 \right)$.
Students should take care that we have put $\left( {{\omega }^{2}}+\omega \right)$ in place of -1. Students should note that, for factored expression all the terms should be in product form. Students can simplify expression as $\left( {{a}^{2}}-ab+{{b}^{2}} \right)={{\left( a-b \right)}^{2}}+ab$ and say it as factored form but this is wrong since there is sum of two terms ${{\left( a-b \right)}^{2}}$ but we need products only.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE