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3+\sqrt[3]{-8}+\left(\sqrt{3}-1\right)^{0}
The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -3 is 3.
3-2+\left(\sqrt{3}-1\right)^{0}
Calculate \sqrt[3]{-8} and get -2.
1+\left(\sqrt{3}-1\right)^{0}
Subtract 2 from 3 to get 1.
1+1
Calculate \sqrt{3}-1 to the power of 0 and get 1.
2
Add 1 and 1 to get 2.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}