Evaluate: `("lim")_(xvecoo)((x+1(x+2(x+3)3-x)`

Explanation

`L=underset(xtooo)lim(root(3)((x+1)(x+2)(x+3))-x)` ltBRgt Using`a-b=(a^(3)-b^(3))/(a^(2)+ab+b^(2)),` we get
`L=underset(xtooo)lim((x+1)(x+2)(x+3)-x^(3))/([(x+1)(x+2)(x+3)]^(2//3)+x^(2)+x[(x+1)(x+2)(x+3)^(1//3)])`
`=underset(xtooo)lim(x^(2)[6+(11)/(x)+(6)/(x^(2))])/(x^(2)[(1+(6)/(x)+(11)/(x^(2))+(6)/x^(2))^(2//3)+1+(1+(6)/(x)+(11)/(x^(2))+(6)/x^(3))^(1//3)]]`
`=(6)/(3)=2`