g) \( \frac{3}{(x-2)}+\frac{3}{x+2}=\frac{2}{x^{2}-1} \)

To solve the equation \( \frac{3}{(x-2)}+\frac{3}{(x+2)}=\frac{2}{(x^{2}-1)} \), we first notice that \( x^2 - 1 \) can be factored as \( (x-1)(x+1) \). Let's rewrite the equation with a common denominator: The common denominator for \( (x-2)(x+2) \) and \( (x-1)(x+1) \) is \( (x-2)(x+2)(x-1)(x+1) \). Multiplying through by the common denominator, we get: \[ 3(x+2)(x-1)(x+1) + 3(x-2)(x-1)(x+1) = 2(x-2)(x+2) \] Now we'll expand everything step by step. 1. Expanding \( 3(x+2)(x-1)(x+1) \): \[ (x-1)(x+1) = x^2 - 1 \] Hence, \[ 3(x+2)(x^2-1) = 3x^3 + 6x - 3x - 6 = 3x^3 + 3x - 6 \] 2. Expanding \( 3(x-2)(x-1)(x+1) \): Again, \[ (x-1)(x+1) = x^2 - 1 \] Hence, \[ 3(x-2)(x^2-1) = 3x^3 - 6x - 3x + 6 = 3x^3 - 9x + 6 \] 3. The left side of the original equation will then be: \[ 3x^3 + 3x - 6 + 3x^3 - 9x + 6 = 6x^3 - 6x \] 4. Now, looking at the right side, \( 2(x-2)(x+2) \): \[ 2(x^2 - 4) = 2x^2 - 8 \] Combining everything gives us: \[ 6x^3 - 6x = 2x^2 - 8 \] Moving all terms to one side: \[ 6x^3 - 2x^2 - 6x + 8 = 0 \] Now, we can factor or use numerical methods or the Rational Root Theorem to find solutions. Testing some simple rational roots (like 1 or -1): Testing \( x = 2 \): \[ 6(2)^3 - 2(2)^2 - 6(2) + 8 = 48 - 8 - 12 + 8 = 36 \text{ (not a root)} \] Testing \( x = 1 \): \[ 6(1)^3 - 2(1)^2 - 6(1) + 8 = 6 - 2 - 6 + 8 = 6 \text{ (not a root)} \] Testing \( x = -1 \): \[ 6(-1)^3 - 2(-1)^2 - 6(-1) + 8 = -6 - 2 + 6 + 8 = 6 \text{ (not a root)} \] Continuing this process or finding another way to factor might help lead to solutions. Ultimately from these contexts, you would simplify, gather terms, and use algebraic techniques to find the values of \( x \). To proceed to solve this polynomial we'd likely need to graph or further find roots.