Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= x^{3} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- x_{n}^{3} + \sin{\left(x_{n} \right)} + 8}{- 3 x_{n}^{2} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- (3.0000000000)^{3} + \sin{\left((3.0000000000) \right)} + 8}{- 3 (3.0000000000)^{2} + \cos{\left((3.0000000000) \right)}} = 2.3262277582$ $LaTeX: x_{2} = (2.3262277582) - \frac{- (2.3262277582)^{3} + \sin{\left((2.3262277582) \right)} + 8}{- 3 (2.3262277582)^{2} + \cos{\left((2.3262277582) \right)}} = 2.0980887335$ $LaTeX: x_{3} = (2.0980887335) - \frac{- (2.0980887335)^{3} + \sin{\left((2.0980887335) \right)} + 8}{- 3 (2.0980887335)^{2} + \cos{\left((2.0980887335) \right)}} = 2.0709853054$ $LaTeX: x_{4} = (2.0709853054) - \frac{- (2.0709853054)^{3} + \sin{\left((2.0709853054) \right)} + 8}{- 3 (2.0709853054)^{2} + \cos{\left((2.0709853054) \right)}} = 2.0706164538$ $LaTeX: x_{5} = (2.0706164538) - \frac{- (2.0706164538)^{3} + \sin{\left((2.0706164538) \right)} + 8}{- 3 (2.0706164538)^{2} + \cos{\left((2.0706164538) \right)}} = 2.0706163859$