Solve (-2)^3-1/3div5*|1-(-4)^2|=

$$( - 2 ) ^ { 3 } - \frac { 1 } { 3 } \div 5 \times | 1 - ( - 4 ) ^ { 2 } | =$$

Evaluate

$$-9$$

Solution Steps

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

$$-8-\frac{\frac{1}{3}}{5}|1-\left(-4\right)^{2}|$$

Express \frac{\frac{1}{3}}{5} as a single fraction.

$$-8-\frac{1}{3\times 5}|1-\left(-4\right)^{2}|$$

Multiply 3 and 5 to get 15.

$$-8-\frac{1}{15}|1-\left(-4\right)^{2}|$$

Calculate -4 to the power of 2 and get 16.

$$-8-\frac{1}{15}|1-16|$$

Subtract 16 from 1 to get -15.

$$-8-\frac{1}{15}|-15|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -15 is 15.

$$-8-\frac{1}{15}\times 15$$

Cancel out 15 and 15.

$$-8-1$$

Subtract 1 from -8 to get -9.

$$-9$$

Factor

$$-9$$