Question

How do you simplify $\dfrac{{{a}^{3}}-{{b}^{3}}}{3{{a}^{2}}+9ab+6{{b}^{2}}}.\dfrac{{{a}^{2}}+2ab+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$ ?

Answer 1

Hint: We have been given a complex polynomial expression where two fractions are multiplied and each term in the numerator and denominator is a 2-degree or 3-degree polynomial. We shall use the various algebraic properties to expand the numerator of the first fraction and the denominator of the second fraction. Also we shall modify the remaining terms to cancel common terms from the entire expression and bring it to its simplest form.

Complete step by step solution:
Given that $\dfrac{{{a}^{3}}-{{b}^{3}}}{3{{a}^{2}}+9ab+6{{b}^{2}}}.\dfrac{{{a}^{2}}+2ab+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$.
We know that according to the fundamental algebraic properties, the difference of the cubes of two variables or constants is expanded as, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, the difference of the squares of two variables or constants is expanded equal to the product of their respective sum and difference as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and the square of the sum of two terms is expanded as, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Applying these three properties simultaneously, we modify the given expression as$\dfrac{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3{{a}^{2}}+9ab+6{{b}^{2}}}.\dfrac{{{\left( a+b \right)}^{2}}}{\left( a+b \right)\left( a-b \right)}$
Cancelling $\left( a+b \right)$ and $\left( a-b \right)$ from the numerator and denominator, we get
$\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3{{a}^{2}}+9ab+6{{b}^{2}}}.\dfrac{\left( a+b \right)}{1}$
Now, we break the middle term $9ab$ into $3ab+6ab$and expand the polynomial $3{{a}^{2}}+9ab+6{{b}^{2}}$
$\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3{{a}^{2}}+3ab+6ab+6{{b}^{2}}}.\dfrac{\left( a+b \right)}{1}$
Here, we will take the repetitive terms common and group the terms.
 $\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3a\left( a+b \right)+6b\left( a+b \right)}.\dfrac{\left( a+b \right)}{1}$
$\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{\left( a+b \right)\left( 3a+6b \right)}.\dfrac{\left( a+b \right)}{1}$
Again, cancelling the common term $\left( a+b \right)$, we get
 $\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{\left( 3a+6b \right)}.\dfrac{1}{1}$
Also, we shall take 3 common in the denominator because $3\times 2=6$.
$\Rightarrow \dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3\left( a+2b \right)}$

Therefore, $\dfrac{{{a}^{3}}-{{b}^{3}}}{3{{a}^{2}}+9ab+6{{b}^{2}}}.\dfrac{{{a}^{2}}+2ab+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$ is simplified to $\dfrac{\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{3\left( a+2b \right)}$.

Note: In order to simplify and solve expressions consisting of multiple polynomials, we must have prior knowledge of the fundamental algebraic properties. One mistake that we could have made while solving this question was that we could have used a negative sign in the middle term instead of positive while expanding ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$which would change the expression to ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and make the solution completely incorrect.