Solve [-1-(-1/4)^{2}]div[1+(-1/4)^2]*[(-2)^3+(-3)^2]^2

$$[ - 1 - ( - \frac { 1 } { 4 } ) ^ { 2 } ] \div [ 1 + ( - \frac { 1 } { 4 } ) ^ { 2 } ] \times [ ( - 2 ) ^ { 3 } + ( - 3 ) ^ { 2 } ] ^ { 2 }$$

Evaluate

$$-1$$

Solution Steps

Extract the negative sign in -1-\left(-\frac{1}{4}\right)^{2}.

$$\frac{-\left(1+\left(-\frac{1}{4}\right)^{2}\right)}{1+\left(-\frac{1}{4}\right)^{2}}\left(\left(-2\right)^{3}+\left(-3\right)^{2}\right)^{2}$$

Cancel out 1+\left(-\frac{1}{4}\right)^{2} in both numerator and denominator.

$$-\left(\left(-2\right)^{3}+\left(-3\right)^{2}\right)^{2}$$

Calculate -2 to the power of 3 and get -8.

$$-\left(-8+\left(-3\right)^{2}\right)^{2}$$

Calculate -3 to the power of 2 and get 9.

$$-\left(-8+9\right)^{2}$$

Add -8 and 9 to get 1.

$$-1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$-1$$

Factor

$$-1$$