Find the roots :
`(x-3)/(x+3) -(x+3)/(x-3)=0,(x != 3,-3)`

Answer

To find the roots of the equation

\[ \frac{x-3}{x+3} - \frac{x+3}{x-3} = 0, \quad (x \neq 3, -3) \]

we will follow these steps:

### Step 1: Take the LCM
The least common multiple (LCM) of the denominators \(x + 3\) and \(x - 3\) is \((x + 3)(x - 3)\).

So, we can rewrite the equation as:

\[ \frac{(x-3)^2 - (x+3)^2}{(x+3)(x-3)} = 0 \]

### Step 2: Set the Numerator to Zero
For the fraction to be equal to zero, the numerator must be zero. Therefore, we set:

\[ (x-3)^2 - (x+3)^2 = 0 \]

### Step 3: Apply the Difference of Squares Formula
The expression \((a^2 - b^2)\) can be factored using the formula \(a^2 - b^2 = (a + b)(a - b)\). Here, let \(a = (x-3)\) and \(b = (x+3)\):

\[ [(x-3) + (x+3)][(x-3) - (x+3)] = 0 \]

### Step 4: Simplify the Factors
Now, simplify each factor:

1. \((x-3) + (x+3) = 2x\)
2. \((x-3) - (x+3) = -6\)

Thus, we have:

\[ (2x)(-6) = 0 \]

### Step 5: Solve for \(x\)
Setting each factor equal to zero gives us:

1. \(2x = 0 \Rightarrow x = 0\)
2. \(-6 = 0\) (This does not provide a solution)

Thus, the only solution is:

\[ x = 0 \]

### Step 6: Verify the Solution
To verify, substitute \(x = 0\) back into the original equation:

\[ \frac{0-3}{0+3} - \frac{0+3}{0-3} = \frac{-3}{3} - \frac{3}{-3} = -1 + 1 = 0 \]

Since the left-hand side equals zero, our solution is confirmed.

### Final Answer
The root of the equation is:

\[ x = 0 \]

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