Solve (-1)^2003-|-2|+(sqrt{3}-x)^0*sqrt[3]{8}+(1/4)^-1

$$( - 1 ) ^ { 2003 } - | - 2 | + ( \sqrt { 3 } - x ) ^ { 0 } \times \sqrt[ 3 ] { 8 } + ( \frac { 1 } { 4 } ) ^ { - 1 }$$

Evaluate

$$3$$

Solution Steps

Calculate -1 to the power of 2003 and get -1.

$$-1-|-2|+\left(\sqrt{3}-x\right)^{0}\sqrt[3]{8}+\left(\frac{1}{4}\right)^{-1}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$-1-2+\left(\sqrt{3}-x\right)^{0}\sqrt[3]{8}+\left(\frac{1}{4}\right)^{-1}$$

Subtract 2 from -1 to get -3.

$$-3+\left(\sqrt{3}-x\right)^{0}\sqrt[3]{8}+\left(\frac{1}{4}\right)^{-1}$$

Calculate \sqrt{3}-x to the power of 0 and get 1.

$$-3+1\sqrt[3]{8}+\left(\frac{1}{4}\right)^{-1}$$

Calculate \sqrt[3]{8} and get 2.

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

Multiply 1 and 2 to get 2.

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

Add -3 and 2 to get -1.

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

Calculate \frac{1}{4} to the power of -1 and get 4.

$$-1+4$$

Add -1 and 4 to get 3.

$$3$$

Factor

$$3$$