Resolve the following expression ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into factors.
Answer
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Hint: We can write the given expression ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ as ${{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$. Then we can use the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$ to split ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into its factors.
Complete step-by-step answer:
Before proceeding with the question, we must know what is meant by factorization. Factorization is defined as the breaking of an entity like a number, a polynomial, etc, into a product of another entity, or factors, which when multiplied together give the original number. We must know the formula of ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ and ${{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac \right)$.
In the question, we have been given to resolve ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into factors.
Now, first, we need to know the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$. For that, we can replace ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ in ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ as shown below,
$\Rightarrow {{\left( a+b \right)}^{3}}-3ab\left( a+b \right)+{{c}^{3}}-3abc$
Collecting the cubic terms together we get:
$\Rightarrow {{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right)-3abc$
Now, applying the formula of ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$, we get:
$\Rightarrow {{\left( a+b+c \right)}^{3}}-3\left( a+b \right)\left( c \right)\left( a+b+c \right)-3ab\left( a+b \right)-3abc$
Taking $3ab$ as common from the above equation we get:
$\Rightarrow {{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b+c \right)$
Taking out $\left( a+b+c \right)$ as common we get:
$\Rightarrow \left( a+b+c \right)\left[ {{\left( a+b+c \right)}^{2}}-3\left( a+b \right)c-3ab \right]$
Now, we can split ${{\left( a+b+c \right)}^{2}}$ using the formula ${{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac \right)$ and we get:
$\Rightarrow \left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac-3ac-3bc-3ab \right]$
$\Rightarrow \left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right]$
Therefore, we have got that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$.
Now as we have to factorize ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$, we have to replace ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ with ${{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$, so that it could be in the form of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$, where we can use the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$.
$\therefore {{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc={{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$
Now using the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$, we get:
$\Rightarrow {{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c=\left( a-b+c \right)\left( {{a}^{2}}+{{\left( -b \right)}^{2}}+{{c}^{2}}-a\left( -b \right)-\left( -b \right)c-ca \right)$
$\Rightarrow {{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc=\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$
Therefore, after resolving ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into factors we get: $\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$
Hence, the answer is $\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$.
Note: We must not forget to replace ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ with ${{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$, because only then it will be in the form of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ and we can apply the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$ without having any confusion regarding the signs. In the problems of factorization, we must try to take the terms common wherever possible. We must try to simplify the problem before applying all the formulas related to the problems. Remember that the formula ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ can also be written in the form of ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$, therefore, we must be careful with these forms because some questions can only be solved using the first form.
Complete step-by-step answer:
Before proceeding with the question, we must know what is meant by factorization. Factorization is defined as the breaking of an entity like a number, a polynomial, etc, into a product of another entity, or factors, which when multiplied together give the original number. We must know the formula of ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ and ${{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac \right)$.
In the question, we have been given to resolve ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into factors.
Now, first, we need to know the value of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$. For that, we can replace ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ in ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ as shown below,
$\Rightarrow {{\left( a+b \right)}^{3}}-3ab\left( a+b \right)+{{c}^{3}}-3abc$
Collecting the cubic terms together we get:
$\Rightarrow {{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right)-3abc$
Now, applying the formula of ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$, we get:
$\Rightarrow {{\left( a+b+c \right)}^{3}}-3\left( a+b \right)\left( c \right)\left( a+b+c \right)-3ab\left( a+b \right)-3abc$
Taking $3ab$ as common from the above equation we get:
$\Rightarrow {{\left( a+b+c \right)}^{3}}-3\left( a+b \right)c\left( a+b+c \right)-3ab\left( a+b+c \right)$
Taking out $\left( a+b+c \right)$ as common we get:
$\Rightarrow \left( a+b+c \right)\left[ {{\left( a+b+c \right)}^{2}}-3\left( a+b \right)c-3ab \right]$
Now, we can split ${{\left( a+b+c \right)}^{2}}$ using the formula ${{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac \right)$ and we get:
$\Rightarrow \left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac-3ac-3bc-3ab \right]$
$\Rightarrow \left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right]$
Therefore, we have got that ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$.
Now as we have to factorize ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$, we have to replace ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ with ${{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$, so that it could be in the form of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$, where we can use the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$.
$\therefore {{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc={{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$
Now using the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$, we get:
$\Rightarrow {{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c=\left( a-b+c \right)\left( {{a}^{2}}+{{\left( -b \right)}^{2}}+{{c}^{2}}-a\left( -b \right)-\left( -b \right)c-ca \right)$
$\Rightarrow {{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc=\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$
Therefore, after resolving ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ into factors we get: $\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$
Hence, the answer is $\left( a-b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc-ac \right)$.
Note: We must not forget to replace ${{a}^{3}}-{{b}^{3}}+{{c}^{3}}+3abc$ with ${{a}^{3}}+{{\left( -b \right)}^{3}}+{{c}^{3}}-3a\left( -b \right)c$, because only then it will be in the form of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ and we can apply the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$ without having any confusion regarding the signs. In the problems of factorization, we must try to take the terms common wherever possible. We must try to simplify the problem before applying all the formulas related to the problems. Remember that the formula ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ can also be written in the form of ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$, therefore, we must be careful with these forms because some questions can only be solved using the first form.
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