Solve (x+3)(x-3)+(x+1)(x-1)=0: Perfect Square Differences

To solve the equation $(x+3)(x-3) + (x+1)(x-1) = 0$, we will employ the difference of squares formula.

Step 1: Simplify $(x+3)(x-3)$ using the difference of squares:
$(x+3)(x-3) = x^2 - 3^2 = x^2 - 9$.

Step 2: Simplify $(x+1)(x-1)$ using the difference of squares:
$(x+1)(x-1) = x^2 - 1^2 = x^2 - 1$.

Step 3: Substitute the simplified expressions back into the original equation:
$x^2 - 9 + x^2 - 1 = 0$.

Step 4: Combine like terms:
$2x^2 - 10 = 0$.

Step 5: Simplify the equation by factoring or isolating $x^2$:
Divide through by 2 to get $x^2 - 5 = 0$.

Step 6: Solve for $x^2$:
$x^2 = 5$.

Step 7: Solve for $x$ by taking the square root of both sides:
$x = \pm \sqrt{5}$.

Therefore, the solution to the equation is $x = \pm \sqrt{5}$.