If 3x = a + b + c, Find the Value of (x − a)³ + (x − b)³ + (x − c)³ − 3(x − a)(x − b)(x − c)
If \[ 3x=a+b+c, \] then the value of \[ (x-a)^3+(x-b)^3+(x-c)^3 -3(x-a)(x-b)(x-c) \] is
(a) \(a+b+c\)
(b) \((a-b)(b-c)(c-a)\)
(c) \(0\)
(d) none of these
Solution
Let \[ p=x-a,\quad q=x-b,\quad r=x-c \]
Then,
\[ p+q+r =(x-a)+(x-b)+(x-c) \]
\[ =3x-(a+b+c) \]
Since \[ 3x=a+b+c \]
\[ p+q+r=0 \]
Using identity:
\[ p^3+q^3+r^3-3pqr =(p+q+r)(p^2+q^2+r^2-pq-qr-rp) \]
\[ =0 \]
Therefore,
\[ \boxed{(c)\ 0} \]