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**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.

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