1. **State the problem:** Simplify the expression $$\left(\frac{y+3}{y^2+3y+9}\right) + \left(\frac{y-3}{y^2-3y+9}\right) - \left(\frac{54}{y^4+9y^2+81}\right).$$ 2. **Factor the denominators where possible:** - Note that $$y^2+3y+9$$ and $$y^2-3y+9$$ are quadratic expressions that do not factor nicely over the reals. - The denominator $$y^4+9y^2+81$$ can be rewritten as $$\left(y^2\right)^2 + 9y^2 + 81$$. 3. **Rewrite the last denominator:** $$y^4+9y^2+81 = (y^2)^2 + 9y^2 + 81 = y^4 + 9y^2 + 81.$$ This is a quartic polynomial. 4. **Try to factor the quartic:** We attempt to factor $$y^4 + 9y^2 + 81$$ as a product of two quadratics: $$ (y^2 + ay + b)(y^2 + cy + d) = y^4 + (a+c)y^3 + (ac + b + d)y^2 + (ad + bc)y + bd. $$ Matching coefficients with $$y^4 + 0y^3 + 9y^2 + 0y + 81$$ gives: - $$a + c = 0$$ - $$ac + b + d = 9$$ - $$ad + bc = 0$$ - $$bd = 81$$ Try $$a = 3$$ and $$c = -3$$ (since $$a + c = 0$$). Then: - $$ac + b + d = 3 \times (-3) + b + d = -9 + b + d = 9 \Rightarrow b + d = 18$$ - $$ad + bc = 3d + (-3)b = 0 \Rightarrow 3d = 3b \Rightarrow d = b$$ - $$bd = b \times d = b^2 = 81 \Rightarrow b = 9$$ (taking positive root for simplicity) So $$b = d = 9$$. Therefore, $$y^4 + 9y^2 + 81 = (y^2 + 3y + 9)(y^2 - 3y + 9).$$ 5. **Rewrite the original expression using this factorization:** $$\left(\frac{y+3}{y^2+3y+9}\right) + \left(\frac{y-3}{y^2-3y+9}\right) - \left(\frac{54}{(y^2+3y+9)(y^2-3y+9)}\right).$$ 6. **Find common denominator:** The common denominator is $$ (y^2+3y+9)(y^2-3y+9). $$ 7. **Rewrite each term with the common denominator:** - First term numerator: $$ (y+3)(y^2 - 3y + 9) $$ - Second term numerator: $$ (y-3)(y^2 + 3y + 9) $$ - Third term numerator: $$ 54 $$ 8. **Expand the numerators:** - $$ (y+3)(y^2 - 3y + 9) = y^3 - 3y^2 + 9y + 3y^2 - 9y + 27 = y^3 + 0y^2 + 0y + 27 = y^3 + 27 $$ - $$ (y-3)(y^2 + 3y + 9) = y^3 + 3y^2 + 9y - 3y^2 - 9y - 27 = y^3 + 0y^2 + 0y - 27 = y^3 - 27 $$ 9. **Sum the numerators:** $$ (y^3 + 27) + (y^3 - 27) - 54 = y^3 + 27 + y^3 - 27 - 54 = 2y^3 - 54. $$ 10. **Factor numerator:** $$ 2y^3 - 54 = 2(y^3 - 27) = 2(y - 3)(y^2 + 3y + 9). $$ 11. **Write the full expression:** $$ \frac{2(y - 3)(y^2 + 3y + 9)}{(y^2 + 3y + 9)(y^2 - 3y + 9)}. $$ 12. **Cancel common factor $$y^2 + 3y + 9$$:** $$ \frac{2(y - 3)\cancel{(y^2 + 3y + 9)}}{\cancel{(y^2 + 3y + 9)}(y^2 - 3y + 9)} = \frac{2(y - 3)}{y^2 - 3y + 9}. $$ **Final answer:** $$\boxed{\frac{2(y - 3)}{y^2 - 3y + 9}}.$$