Question

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

Hint:

This problem is a direct application of a common algebraic identity. Recognizing expressions like $A^3+B^3+C^3-3ABC$ should prompt you to check if the sum $A+B+C$ equals zero, as this simplifies the expression considerably.

Answer 1

The Correct Option is C

Step 1: Recall the algebraic identity for sum of cubes.
A key algebraic identity states: If $A + B + C = 0$, then $A^3 + B^3 + C^3 = 3ABC$.
This can be rearranged as: $A^3 + B^3 + C^3 - 3ABC = 0$. Step 2: Define A, B, and C in terms of the given expression.
Let's assign the terms inside the cubes to variables:
Let $A = (x-a)$
Let $B = (x-b)$
Let $C = (x-c)$
The expression we need to evaluate is exactly in the form $[A^3 + B^3 + C^3 - 3ABC]$.
Step 3: Check if the condition $A + B + C = 0$ is met.
Calculate the sum of A, B, and C:
$A + B + C = (x-a) + (x-b) + (x-c)$
$A + B + C = x - a + x - b + x - c$
$A + B + C = 3x - (a + b + c)$
Step 4: Use the given information to simplify $A + B + C$.
The problem states that $3x = a + b + c$.
Substitute this into the sum $A + B + C$:
$A + B + C = (a + b + c) - (a + b + c)$
$A + B + C = 0$
Step 5: Apply the algebraic identity.
Since $A + B + C = 0$, according to the identity mentioned in Step 1: $A^3 + B^3 + C^3 - 3ABC = 0$
Therefore, the value of the given expression is 0.
Step 6: Compare the result with the given options.
The calculated value is 0, which matches option (3). $$(3) 0$$