Solve -1^3-3*-1+2=1-3+2=0

$${ -1 }^{ 3 } -3 \cdot -1+2 = 1-3+2=0$$

Verify

Solution Steps

Calculate -1 to the power of 3 and get -1.

$$-1-3\left(-1\right)+2=1-3+2\text{ and }1-3+2=0$$

Multiply 3 and -1 to get -3.

$$-1-\left(-3\right)+2=1-3+2\text{ and }1-3+2=0$$

The opposite of -3 is 3.

$$-1+3+2=1-3+2\text{ and }1-3+2=0$$

Add -1 and 3 to get 2.

$$2+2=1-3+2\text{ and }1-3+2=0$$

Add 2 and 2 to get 4.

$$4=1-3+2\text{ and }1-3+2=0$$

Subtract 3 from 1 to get -2.

$$4=-2+2\text{ and }1-3+2=0$$

Add -2 and 2 to get 0.

$$4=0\text{ and }1-3+2=0$$

Compare 4 and 0.

$$\text{false}\text{ and }1-3+2=0$$

Subtract 3 from 1 to get -2.

$$\text{false}\text{ and }-2+2=0$$

Add -2 and 2 to get 0.

$$\text{false}\text{ and }0=0$$

Compare 0 and 0.

$$\text{false}\text{ and }\text{true}$$

The conjunction of \text{false} and \text{true} is \text{false}.

$$\text{false}$$