Solve the equation: \( w^{3}+9 w^{2}-9 w-81=0 \) Enter any solution as an integer or reduced fraction (not a mixed number or decimal), \( u \) there is more than one solution. If there is no solutions click No solution. Suggestion: Ch in the original equation before submitting answer.

To solve the equation \( w^{3}+9 w^{2}-9 w-81=0 \), we can use the Rational Root Theorem to find potential rational roots. Testing \( w = 3 \), we find: \[ 3^3 + 9(3^2) - 9(3) - 81 = 27 + 81 - 27 - 81 = 0 \] Thus, \( w = 3 \) is a root. Now we can factor the polynomial using synthetic division or polynomial long division: Dividing \( w^3 + 9w^2 - 9w - 81 \) by \( w - 3 \) gives us \( w^2 + 12w + 27 \). Next, we can solve \( w^2 + 12w + 27 = 0 \) using the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 108}}{2} = \frac{-12 \pm \sqrt{36}}{2} \] This simplifies to: \[ w = \frac{-12 \pm 6}{2} \] Calculating the two solutions: 1. \( w = \frac{-6}{2} = -3 \) 2. \( w = \frac{-18}{2} = -9 \) Thus, the three solutions to the equation \( w^{3}+9 w^{2}-9 w-81=0 \) are: \[ w = 3, \quad w = -3, \quad w = -9 \] In this case, entering any solution such as \( 3 \) would be valid.