Find the value of x if: `(x-4)^3+(x-9)^3+(x-8)^3=3(x-4)(x-9)(x-8)`

Answer

To solve the equation \((x-4)^3 + (x-9)^3 + (x-8)^3 = 3(x-4)(x-9)(x-8)\), we can use the identity for the sum of cubes. The identity states that:

\[ a^3 + b^3 + c^3 = 3abc \quad \text{if and only if} \quad a + b + c = 0 \]

### Step 1: Identify \(a\), \(b\), and \(c\)

Let:
- \(a = x - 4\)
- \(b = x - 9\)
- \(c = x - 8\)

### Step 2: Apply the identity

According to the identity, if \(a^3 + b^3 + c^3 = 3abc\), then we must have:

\[ a + b + c = 0 \]

### Step 3: Calculate \(a + b + c\)

Now, we can calculate:

\[ a + b + c = (x - 4) + (x - 9) + (x - 8) \]

Combining the terms:

\[ a + b + c = x - 4 + x - 9 + x - 8 = 3x - 21 \]

### Step 4: Set the equation to zero

Setting \(a + b + c\) equal to zero gives us:

\[ 3x - 21 = 0 \]

### Step 5: Solve for \(x\)

Now, we can solve for \(x\):

\[ 3x = 21 \]

Dividing both sides by 3:

\[ x = \frac{21}{3} = 7 \]

### Final Answer

Thus, the value of \(x\) is:

\[ \boxed{7} \]