3 Perfoun the operation sand operatorname(lif)_(2) (x^2-81)/(x^2)+18 times 81 div ((x-3)(x+1))/(x^2)-6 x+9

To perform the operation \( \operatorname{lif}_{2} \frac{x^{2}-81}{x^{2}+18 \times 81} \div \frac{(x-3)(x+1)}{x^{2}-6 x+9} \), we need to follow the steps below:Step 1: Simplify the expression inside the fraction.The numerator can be factored as \( (x-9)(x+9) \).The denominator can be simplified as .So, the expression becomes \( \frac{(x-9)(x+9)}{x^{2}+1458} \).Step 2: Simplify the denominator of the second fraction.The denominator can be factored as \( (x-3)^{2} \).So, the expression becomes \( \frac{(x-3)(x+1)}{(x-3)^{2}} \).Step 3: Rewrite the division as multiplication by the reciprocal.The expression becomes \( \frac{(x-9)(x+9)}{x^{2}+1458} \times \frac{(x-3)^{2}}{(x-3)(x+1)} \).Step 4: Cancel out common factors.The factor \( (x-3) \) in the numerator and denominator can be canceled out.So, the expression becomes \( \frac{(x-9)(x+9)}{x^{2}+1458} \times \frac{(x-3)}{(x+1)} \).Step 5: Multiply the fractions.The expression becomes \( \frac{(x-9)(x+9)(x-3)}{(x^{2}+1458)(x+1)} \).Therefore, the final answer is \( \frac{(x-9)(x+9)(x-3)}{(x^{2}+1458)(x+1)} \).