`(i) (-3, 0) = {x : x ∈ R, -3 < x < 0}`

`sum_(i=1)^n i^3=((n(n+1))/2)^2`

\[B(m,n)=\int\limits_{0}^{1}{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}dx}\]