Given expression is (x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)
We know that,
(a+b+c)³ = a³ + b³ + c³ + 3(a+b)(b+c)(c+a) ------------(i)
Expanding (a+b+c)³, we get
a³ + b³ + c³ + 3(a²b+a²c+b²a+b²c+c²a+c²b) + 6abc
Given a+b+c=3x, substituting in above equation, we get
a³ + b³ + c³ + 9(ab+bc+ca) - 6abc = 27x³ ------------(ii)
Now, expanding given expression, we get
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)
x³ - 3x²(a+b+c) + 3x(ab+bc+ca) - (a³+b³+c³-3abc)
Given a+b+c=3x, substituting in above equation, we get
x³ - 9x³ + 9x³ - (a³+b³+c³-3abc)
= -8x³ + (a³+b³+c³+3abc)
Now, using equation (ii), we get
-8x³ + 27x³ - 3(a³+b³+c³-3abc)
= 19x³ + 9abc
Therefore, the given expression is equal to 19x³ + 9abc.
Given a+b+c=3x, we have abc ≤ x³. Hence, the minimum value of 9abc is 9x³.
Substituting this value in above equation, we get
19x³ + 9abc ≥ 19x³ + 9x³ = 28x³
Therefore, the minimum value of given expression is 28x³.
Hence, the correct answer is d) 0.