#### Answer

$$
x^{3}=\tan^{-1} x
$$
The roots of equation, correct to six decimal places, are 0, -0.902025 and 0.902025.

#### Work Step by Step

$$
x^{3}=\tan^{-1} x
$$
We first rewrite the equation in standard form:
$$
x^{3}-\tan^{-1} x=0
$$
Therefore we let
$$
f(x)=x^{3}-\tan^{-1} x=0 .
$$
Then
$$
f^{'}(x)=3x^{2}-\frac{1}{1+ x^{2}}
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{x_{n}^{3}-\tan^{-1} x_{n}}{3x_{n}^{2}-\frac{1}{1+ x_{n}^{2}}}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=x^{3} $ and $y = \tan^{-1} x $ in the Figure .
(*) It appears that they intersect at a point whose x-coordinate is somewhat less than 1, so let’s take $x_{1} = 1$ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{3}-\tan^{-1} x_{1}}{3x_{1}^{2}-\frac{1}{1+ x_{1}^{2}}}\\
&=(1)-\frac{(1)^{3}-\tan^{-1} (1)}{3(1)^{2}-\frac{1}{1+ (1)^{2}}} \approx 0.914159
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 0.902251,
x_{4}\approx 0.902026 ,
x_{5}\approx 0.902025 .
$$
(**) It appears that they intersect at a point whose x-coordinate is somewhat greater than -1, so let’s take $x_{1} = -1 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{3}-\tan^{-1} x_{1}}{3x_{1}^{2}-\frac{1}{1+ x_{1}^{2}}}\\
&=(-1)-\frac{(-1)^{3}-\tan^{-1} (-1)}{3(-1)^{2}-\frac{1}{1+ (-1)^{2}}} \approx -0.914159
\end{aligned}
$$
repeating we get
$$
x_{3}\approx - 0.902251,
x_{4}\approx -0.902026 ,
x_{5}\approx -0.902025 .
$$
we conclude that the roots of equation, correct to six decimal places, are 0, -0.902025 and 0.902025.