Question

The factors of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] are

Answer 1

Hint: From the given question we have been asked to find the factors of given expression. For answering this question we will use factorization and factor theorem. We will assume that \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] be a function in a and we will use the substitution method and solve the given question using factor theorem and basic mathematical operations like addition and subtraction. So, the solution will be as follows.

Complete step by step solution:
We have been given \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\].
So, as mentioned in the above we will assume that,
\[\Rightarrow f(a)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] be a function in a.
Now, we will use the substitution and we will substitute \[a=-(b+c)\] in the above function. So, we get the reduced as follows.
\[\Rightarrow f(-(b+c))={{\left( -\left( b+c \right) \right)}^{3}}+{{b}^{3}}+{{c}^{3}}+3(b+c)bc\]
Now we will expand this equation using basic mathematical operations like addition and subtraction.
\[\Rightarrow f(-(b+c))={{\left( -\left( b+c \right) \right)}^{3}}+{{\left( b+c \right)}^{3}}\]
\[\Rightarrow f(-(b+c)=0\]
Hence, by factor theorem \[(a+b+c)\] is a factor.
Now we take the given \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] function.
\[\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
We can rewrite this expression as follows.
\[\Rightarrow {{a}^{3}}+\left( {{b}^{3}}+{{c}^{3}} \right)-3abc\]
We can rewrite the above expression as follows by adding and subtracting the term \[-3bc\left( b+c \right)\]
\[\Rightarrow {{a}^{3}}+{{\left( b+c \right)}^{3}}-3abc-3bc\left( b+c \right)\]
We already got \[(a+b+c)\] as a factor. So, we sill substitute \[\left( b+c=-a \right)\] in the above expression.
So, we get the expression reduced as follows.
\[\Rightarrow {{a}^{3}}+{{\left( b+c \right)}^{3}}-3abc-3bc\left( -a \right)\]
\[\Rightarrow {{a}^{3}}+{{\left( b+c \right)}^{3}}-3abc+3bca\]
\[\Rightarrow {{a}^{3}}+{{\left( b+c \right)}^{3}}\]
Now we will use the formula \[\Rightarrow {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)\]
So, here by comparing we get \[\left( b+c=y \right)\]. So, we will substitute them in this formula. Now, we get the equation reduced as follows.
\[\Rightarrow \left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2bc-ab-ac \right)\]
So, the factors are \[\Rightarrow \left( a+b+c \right),\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2bc-ab-ac \right)\]

Note: Students should have good knowledge in the concept of factorisation. Students should be able to do calculations without any errors. Students should know the formulae like,
 \[\Rightarrow {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)\] to solve the question.