Question

How do you solve\[\dfrac{2{{a}^{2}}+ab-{{b}^{2}}}{{{a}^{3}}+{{b}^{3}}}\div \dfrac{2{{a}^{2}}{{b}^{2}}-a{{b}^{3}}}{2{{a}^{2}}-2ab+2{{b}^{2}}}\]?

Answer 1

Hint: In the given question, we have been asked to solve the complex algebraic fraction. In order to solve the given question, first we need to factorize the algebraic fraction wherever possible by using the formulae of sum of cubes, by splitting the middle term of the quadratic equation given in any part of the question and by taking out common factors. Lastly cancel out all the like terms and simplify.

Formula Used: The formulae of sum of cubes which states that\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\].

Complete step-by-step solution:
We have given that,
\[\Rightarrow \dfrac{2{{a}^{2}}+ab-{{b}^{2}}}{{{a}^{3}}+{{b}^{3}}}\div \dfrac{2{{a}^{2}}{{b}^{2}}-a{{b}^{3}}}{2{{a}^{2}}-2ab+2{{b}^{2}}}\]
Factorize the given algebraic fraction:
Using the property of sum of cubes, i.e.
 \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]
Applying the property of sum of cubes, we get
\[\Rightarrow \dfrac{2{{a}^{2}}+ab-{{b}^{2}}}{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}\div \dfrac{2{{a}^{2}}{{b}^{2}}-a{{b}^{3}}}{2{{a}^{2}}-2ab+2{{b}^{2}}}\]
Factorize \[2{{a}^{2}}+ab-{{b}^{2}}\] by splitting the middle term,
Now, we have
\[\Rightarrow 2{{a}^{2}}+ab-{{b}^{2}}\]
\[\Rightarrow 2{{a}^{2}}+2ab-ab-{{b}^{2}}\]
\[\Rightarrow 2a\left( a+b \right)-b\left( a+b \right)\]
Taking out common factor, we get
\[\Rightarrow \left( 2a-b \right)\left( a+b \right)\]
Therefore,
Substitute \[2{{a}^{2}}+ab-{{b}^{2}}=\left( 2a-b \right)\left( a+b \right)\] in the plug-in solved algebraic fraction,
\[\Rightarrow \dfrac{\left( 2a-b \right)\left( a+b \right)}{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}\div \dfrac{2{{a}^{2}}{{b}^{2}}-a{{b}^{3}}}{2{{a}^{2}}-2ab+2{{b}^{2}}}\]
Now, factorize the remaining parts of the above algebraic fraction by taking out the common factor,
We obtain,
\[\Rightarrow \dfrac{\left( 2a-b \right)\left( a+b \right)}{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}\div \dfrac{a{{b}^{2}}\left( 2a-b \right)}{2\left( {{a}^{2}}-ab+{{b}^{2}} \right)}\]
To change the divide sign into multiplication, the fraction after the divide sign multiply by the reciprocal;
\[\Rightarrow \dfrac{\left( 2a-b \right)\left( a+b \right)}{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}\times \dfrac{2\left( {{a}^{2}}-ab+{{b}^{2}} \right)}{a{{b}^{2}}\left( 2a-b \right)}\]
Cancelling out all the like factors given in the above algebraic fraction, we get
\[\Rightarrow \dfrac{2}{a{{b}^{2}}}\]
Therefore,
\[\Rightarrow \dfrac{2{{a}^{2}}+ab-{{b}^{2}}}{{{a}^{3}}+{{b}^{3}}}\div \dfrac{2{{a}^{2}}{{b}^{2}}-a{{b}^{3}}}{2{{a}^{2}}-2ab+2{{b}^{2}}}=\dfrac{2}{a{{b}^{2}}}\]
Hence, it is the required solution.

Note: While solving these types of question, students should have been very careful while writing down the term and corresponding sign of that term. To solve this question, all just we need to factorize the given algebraic expression, for that students need to know about the different methods of factorization such as factorization by splitting the middle term of given perfect trinomial, factorization by taking out common factor and factorization by applying algebraic identities.