Hint: Use long division method to divide ${a^4} + {b^4} + {a^2}{b^2}$ by the given factor ${a^2} + {b^2} + ab$ to find other factor.
Complete step by step answer:
We know that, if we are provided with a factor of a number and we have to find another factor we simply divide the number by the given factor.
For example we know that 3 is one of a factor of 12 and we have to find another factor we simply divide 12 by 3 to know another factor i.e.
$ \dfrac{{12}}{3} = 4 $
Thus, 4 is another factor of 12.
In the similar way we will find the factor of given question
Given that ${a^2} + {b^2} + ab$ is a factor of ${a^4} + {b^4} + {a^2}{b^2}$
Now another factor is determined by long division method:
$ {a^2} + {b^2} + ab\mathop{\left){\vphantom{1\begin{gathered}
{a^4} + {b^4} + {a^2}{b^2}
- {a^2}{\text{ - }}{a^2}{b^2}{\text{ - }}{a^3}b
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
{b^4}{\text{ }} - {a^3}b
{b^4}{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
\_\_\_\_\_\_\_\_\_\_\_\_\_\_
- {a^3}b{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
- {a^3}b{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
{\text{ }}0
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\end{gathered} }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered}
{a^4} + {b^4} + {a^2}{b^2}
- {a^2}{\text{ - }}{a^2}{b^2}{\text{ - }}{a^3}b
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
{b^4}{\text{ }} - {a^3}b
{b^4}{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
\_\_\_\_\_\_\_\_\_\_\_\_\_\_
- {a^3}b{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
- {a^3}b{\text{ }} - a{b^3}{\text{ }} - {a^2}{b^2}
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
{\text{ }}0
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\end{gathered} }}}
\limits^{\displaystyle \,\,\, {{a^2} + {b^2} - ab}} $
Thus the another factor is ${a^2} + {b^2} - ab$
$ \therefore$ The correct option is (b)
Note: - When one factor of the polynomial is given and asked us to find another polynomial we will factorise the given polynomial or use a long division method to find the solution.