Find `(a-b)^(3) + (b-c)^(3) + (c-a)^(3)=` ?

Answer

To solve the expression \((a-b)^{3} + (b-c)^{3} + (c-a)^{3}\), we can use a known algebraic identity.

### Step 1: Identify the variables
Let:
- \( x = a - b \)
- \( y = b - c \)
- \( z = c - a \)

### Step 2: Check the sum \( x + y + z \)
Now, we calculate \( x + y + z \):
\[ x + y + z = (a - b) + (b - c) + (c - a) \]
When we simplify this, we have:
\[ x + y + z = a - b + b - c + c - a = 0 \]

### Step 3: Apply the identity for cubes
Since \( x + y + z = 0 \), we can use the identity:
\[ x^3 + y^3 + z^3 = 3xyz \]
Thus, we can write:
\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \]

### Step 4: Final expression
The final expression for \((a-b)^{3} + (b-c)^{3} + (c-a)^{3}\) is:
\[ 3(a-b)(b-c)(c-a) \]

### Summary
Therefore, the value of \((a-b)^{3} + (b-c)^{3} + (c-a)^{3}\) is:
\[ 3(a-b)(b-c)(c-a) \]

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