Factorize the given term $ {{a}^{3}}-{{b}^{3}}+1+3ab $ ?
Answer
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Hint: We start solving the problem by converting the given $ {{a}^{3}}-{{b}^{3}}+1+3ab $ to a form that resembles $ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr $ . We then expand it by making use of the result $ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr=\left( p+q+r \right)\left( {{p}^{2}}+{{q}^{2}}+{{r}^{2}}-pq-qr-pr \right) $ . We then make the necessary calculations to get the required factors of $ {{a}^{3}}-{{b}^{3}}+1+3ab $ .
Complete step by step answer:
According to the problem, we need to factorize the given term $ {{a}^{3}}-{{b}^{3}}+1+3ab $
Let us first simplify the given $ {{a}^{3}}-{{b}^{3}}+1+3ab $ .
So, we have $ {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right) $
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right)\left( 1 \right) $ ---(1).
We can see that the term $ {{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right)\left( 1 \right) $ resembles the term $ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr $ . We know that this is equal to $ \left( p+q+r \right)\left( {{p}^{2}}+{{q}^{2}}+{{r}^{2}}-pq-qr-pr \right) $ . We use this result in equation (1).
So, we get $ {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{\left( -b \right)}^{2}}+{{1}^{2}}-\left( a \right)\left( -b \right)-\left( -b \right)\left( 1 \right)-\left( 1 \right)\left( a \right) \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab-1 \right) $ .
We can see that the factors of $ {{a}^{3}}-{{b}^{3}}+1+3ab $ are $ a-b+1 $ and $ {{a}^{2}}+{{b}^{2}}-a+b+ab-1 $ .
$ \therefore $ The factorization of $ {{a}^{3}}-{{b}^{3}}+1+3ab $ is $ \left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab-1 \right) $ .
Note:
Whenever we get this type of problem, we try to convert it into the standard forms to make use of their results. We can also solve the given problem as shown below:
We have given $ {{a}^{3}}-{{b}^{3}}+1+3ab $ ---(2).
We know that $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right) $ . Let us use this result in equation in (2).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)+1+3ab $ ---(3).
Let us add and subtract $ 2ab $ to $ {{a}^{2}}+{{b}^{2}}+ab $ in equation (3).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab+2ab-2ab \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+3ab \right)+1+3ab $ ---(4).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . Let us use this in equation (4).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{\left( a-b \right)}^{2}}+3ab \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+3ab\left( a-b \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+{{1}^{3}}+3ab\left( a-b \right)+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+{{1}^{3}}+3ab\left( a-b+1 \right) $ ---(5).
We can see that $ {{\left( a-b \right)}^{3}}+{{1}^{3}} $ resembles $ {{p}^{3}}+{{q}^{3}} $ which is equal to $ \left( p+q \right)\left( {{p}^{2}}+{{q}^{2}}-pq \right) $ . Let us use this result in equation (5).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{\left( a-b \right)}^{2}}+{{1}^{2}}-\left( a-b \right)\left( 1 \right) \right)+3ab\left( a-b+1 \right) $ ---(6).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . Let us use this in equation (6).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+1-a+b \right)+3ab\left( a-b+1 \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+1-a+b+3ab \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab+1 \right) $ .
Complete step by step answer:
According to the problem, we need to factorize the given term $ {{a}^{3}}-{{b}^{3}}+1+3ab $
Let us first simplify the given $ {{a}^{3}}-{{b}^{3}}+1+3ab $ .
So, we have $ {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right) $
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right)\left( 1 \right) $ ---(1).
We can see that the term $ {{\left( a \right)}^{3}}+{{\left( -b \right)}^{3}}+{{\left( 1 \right)}^{3}}-3\left( a \right)\left( -b \right)\left( 1 \right) $ resembles the term $ {{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr $ . We know that this is equal to $ \left( p+q+r \right)\left( {{p}^{2}}+{{q}^{2}}+{{r}^{2}}-pq-qr-pr \right) $ . We use this result in equation (1).
So, we get $ {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{\left( -b \right)}^{2}}+{{1}^{2}}-\left( a \right)\left( -b \right)-\left( -b \right)\left( 1 \right)-\left( 1 \right)\left( a \right) \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab-1 \right) $ .
We can see that the factors of $ {{a}^{3}}-{{b}^{3}}+1+3ab $ are $ a-b+1 $ and $ {{a}^{2}}+{{b}^{2}}-a+b+ab-1 $ .
$ \therefore $ The factorization of $ {{a}^{3}}-{{b}^{3}}+1+3ab $ is $ \left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab-1 \right) $ .
Note:
Whenever we get this type of problem, we try to convert it into the standard forms to make use of their results. We can also solve the given problem as shown below:
We have given $ {{a}^{3}}-{{b}^{3}}+1+3ab $ ---(2).
We know that $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right) $ . Let us use this result in equation in (2).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)+1+3ab $ ---(3).
Let us add and subtract $ 2ab $ to $ {{a}^{2}}+{{b}^{2}}+ab $ in equation (3).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab+2ab-2ab \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+3ab \right)+1+3ab $ ---(4).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . Let us use this in equation (4).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b \right)\left( {{\left( a-b \right)}^{2}}+3ab \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+3ab\left( a-b \right)+1+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+{{1}^{3}}+3ab\left( a-b \right)+3ab $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab={{\left( a-b \right)}^{3}}+{{1}^{3}}+3ab\left( a-b+1 \right) $ ---(5).
We can see that $ {{\left( a-b \right)}^{3}}+{{1}^{3}} $ resembles $ {{p}^{3}}+{{q}^{3}} $ which is equal to $ \left( p+q \right)\left( {{p}^{2}}+{{q}^{2}}-pq \right) $ . Let us use this result in equation (5).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{\left( a-b \right)}^{2}}+{{1}^{2}}-\left( a-b \right)\left( 1 \right) \right)+3ab\left( a-b+1 \right) $ ---(6).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ . Let us use this in equation (6).
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+1-a+b \right)+3ab\left( a-b+1 \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+1-a+b+3ab \right) $ .
$ \Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}-a+b+ab+1 \right) $ .
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