Text Simplify the expression: (1x−3−1x+3)⋅x2−96.\left(\frac{1}{x-3}-\frac{1}{x+3}\right)\cdot\frac{x^{2}-9}{6}.(x−31−x+31)⋅6x2−9. Solution The bracket: (x+3)−(x−3)(x−3)(x+3)=6x2−9\dfrac{(x+3)-(x-3)}{(x-3)(x+3)}=\dfrac{6}{x^{2}-9}(x−3)(x+3)(x+3)−(x−3)=x2−96. Multiply by x2−96\dfrac{x^{2}-9}{6}6x2−9: 6x2−9⋅x2−96=1,x≠±3.\frac{6}{x^{2}-9}\cdot\frac{x^{2}-9}{6}=1,\qquad x\neq\pm 3.x2−96⋅6x2−9=1,x=±3. The expression is constantly equal to 111. 1 \boxed{\,1\,}1