Find the positive real root of x³ - x - 3 = 0 using Newton-Raphson method. If the starting guess (x₀) is 2, the numerical value of the root after two iterations (x₂) is __________ (round off to two decimal places).

The correct answer is 1.67.

To find the positive real root of the equation using the Newton-Raphson method, we start by defining the function and its first derivative:
Let the function be:
$$f(x)=x^3-x-3$$
Taking the derivative with respect to $x$, we get:
$$f^{\prime }(x)=3x^2-1$$

The Newton-Raphson formula for updating the estimate of the root at step $n+1$ is:
$$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$$

Step 1: First Iteration (Calculating $x_1$)
Given the starting guess $x_0=2$:
First, compute $f(x_0)$ and $f^{\prime }(x_0)$:
$f(2)=2^3-2-3=8-2-3=3$
$f^{\prime }(2)=3(2^2)-1=12-1=11$

Now, calculate $x_1$:
$$x_1=2-\frac{3}{11}\approx 2-0.272727=1.72727$$
Rounding to four decimal places, we get $x_1\approx 1.7273$.

Step 2: Second Iteration (Calculating $x_2$)
Using $x_1=1.7273$, we evaluate the function and its derivative:
$f(1.7273)=(1.7273{)}^3-1.7273-3\approx 5.1537-1.7273-3=0.4264$
$f^{\prime }(1.7273)=3(1.7273{)}^2-1\approx 3(2.9836)-1=8.9508-1=7.9508$

Now, calculate $x_2$:
$$x_2=1.7273-\frac{(1.7273{)}^3-1.7273-3}{3(1.7273{)}^2-1}\approx 1.7273-\frac{0.4264}{7.9508}\approx 1.7273-0.05363=1.67367$$

Rounding the value of $x_2$ to two decimal places, we get:
$$x_2\approx 1.67$$