Question

Find the value of \[{{\left( x-a \right)}^{3}}+{{\left( x-b \right)}^{3}}+{{\left( x-c \right)}^{3}}-3\left( x-a \right)\left( x-b \right)\left( x-c \right)\] when \[a+b+c=3x\].

Answer 1

Hint: To solve this problem we have to know the formula of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]. We have to assign values for a, b, c. Substitute in the formula and further solving gives us the final solution.
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)\]

Complete step-by-step answer:
Given \[{{\left( x-a \right)}^{3}}+{{\left( x-b \right)}^{3}}+{{\left( x-c \right)}^{3}}-3\left( x-a \right)\left( x-b \right)\left( x-c \right)\]
We have to find the value when \[a+b+c=3x\].
We know that
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)\] . . . . . . . . . . . . . . . . . . . . . . . (1)
Now considering that,
If \[a+b+c=0\] then \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Now considering that
\[a=x-a\],
\[b=x-b\],
\[c=x-c\]
Substituting the values in (2) we can write,
\[a+b+c\]= \[\left( x-a \right)+\left( x-b \right)+\left( x-c \right)\]
\[\left( x-a \right)+\left( x-b \right)+\left( x-c \right)\]=\[3x-(a+b+c)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
In the question it was given \[a+b+c=3x\]. . . . . . . . . . . . . . . . (b)
Substituting value of (b) in (a) we get,
\[3x-3x=0\]
Thus \[{{\left( x-a \right)}^{3}}+{{\left( x-b \right)}^{3}}+{{\left( x-c \right)}^{3}}-3\left( x-a \right)\left( x-b \right)\left( x-c \right)\]=0

Note: In (2) we have considered \[a+b+c=0\]then the given question can be written in the form of
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]. In (b) the value is given in the question, so the value of the given expression is zero. Take care while doing calculations.