Show that \( 5 a^{2}-9(b-a)+\frac{2}{b^{5}}+7 a b=\frac{-5 a}{b}-6 a^{3}-(a b)^{4}+\frac{9}{b^{2}-a^{3}} \) when \( a=-2 \) and \( b=-1 \).

To verify the equation \[ 5 a^{2}-9(b-a)+\frac{2}{b^{5}}+7 a b=\frac{-5 a}{b}-6 a^{3}-(a b)^{4}+\frac{9}{b^{2}-a^{3}} \] when \( a = -2 \) and \( b = -1 \), we will substitute these values into both sides of the equation. ### Left Side Calculation First, we calculate the left side: \[ 5 a^{2} = 5(-2)^{2} = 5 \times 4 = 20 \] \[ -9(b-a) = -9(-1 - (-2)) = -9(-1 + 2) = -9(1) = -9 \] \[ \frac{2}{b^{5}} = \frac{2}{(-1)^{5}} = \frac{2}{-1} = -2 \] \[ 7 a b = 7(-2)(-1) = 14 \] Now, summing these results: \[ 20 - 9 - 2 + 14 = 20 + 14 - 9 - 2 = 32 - 11 = 21 \] ### Right Side Calculation Now we calculate the right side: \[ \frac{-5 a}{b} = \frac{-5(-2)}{-1} = \frac{10}{-1} = -10 \] \[ -6 a^{3} = -6(-2)^{3} = -6(-8) = 48 \] \[ -(a b)^{4} = -((-2)(-1))^{4} = -2^{4} = -16 \] \[ \frac{9}{b^{2}-a^{3}} = \frac{9}{(-1)^{2} - (-2)^{3}} = \frac{9}{1 - (-8)} = \frac{9}{1 + 8} = \frac{9}{9} = 1 \] Now we sum these results: \[ -10 + 48 - 16 + 1 = 48 - 26 = 22 - 10 = 21 \] ### Conclusion Both sides equal \( 21 \): \[ 5 a^2 - 9(b-a) + \frac{2}{b^5} + 7ab = 21 \] \[ \frac{-5a}{b} - 6a^3 - (ab)^4 + \frac{9}{b^2 - a^3} = 21 \] Thus, the equation holds true when \( a = -2 \) and \( b = -1 \).