Solve sqrt{-2+6-(-2*3+2^2)^0-1(-1)^3}

$$\sqrt { - 2 + 6 - ( - 2 \cdot 3 + 2 ^ { 2 } ) ^ { 0 } - 1 ( - 1 ) ^ { 3 } }$$

Evaluate

$$2$$

Solution Steps

Add -2 and 6 to get 4.

$$\sqrt{4-\left(-2\times 3+2^{2}\right)^{0}-1\left(-1\right)^{3}}$$

Multiply -2 and 3 to get -6.

$$\sqrt{4-\left(-6+2^{2}\right)^{0}-1\left(-1\right)^{3}}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{4-\left(-6+4\right)^{0}-1\left(-1\right)^{3}}$$

Add -6 and 4 to get -2.

$$\sqrt{4-\left(-2\right)^{0}-1\left(-1\right)^{3}}$$

Calculate -2 to the power of 0 and get 1.

$$\sqrt{4-1-1\left(-1\right)^{3}}$$

Subtract 1 from 4 to get 3.

$$\sqrt{3-1\left(-1\right)^{3}}$$

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

$$\sqrt{3-1\left(-1\right)}$$

Multiply 1 and -1 to get -1.

$$\sqrt{3-\left(-1\right)}$$

The opposite of -1 is 1.

$$\sqrt{3+1}$$

Add 3 and 1 to get 4.

$$\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$2$$

Factor

$$2$$