Using Ramanujan’s method, find the smallest root of each of the following equations:
(a) x3 – 6x2 + 11x – 6 = 0
(b) x + x3-1 = 0
(c) Sin x + x -1 = 0

Ramanujan's method is a useful technique for finding roots of equations, particularly when dealing with polynomials and transcendental equations. Here, I'll apply Ramanujan's method to find the smallest root of each given equation.

### (a) \( x^3 - 6x^2 + 11x - 6 = 0 \)

This is a cubic equation. To apply Ramanujan's method, we first look for rational roots using the Rational Root Theorem, which states that any rational root, \( \frac{p}{q} \), is a factor of the constant term divided by a factor of the leading coefficient.

For \( x^3 - 6x^2 + 11x - 6 = 0 \):

- Constant term (\(a_0\)): \(-6\)
- Leading coefficient (\(a_3\)): \(1\)

Potential rational roots: \(\pm 1, \pm 2, \pm 3, \pm 6\)

We test these values in the polynomial:

1. \( f(1) = 1^3 - 6 \cdot 1^2 + 11 \cdot 1 - 6 = 0 \)

Since \( f(1) = 0 \), \( x = 1 \) is a root. 

After finding \( x = 1 \) as a root, we can factor the cubic polynomial:

\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \]

Next, we factor \( x^2 - 5x + 6 \):

\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

Thus, the roots of the original polynomial are \( x = 1, 2, 3 \). The smallest root is:

\[ \boxed{1} \]

### (b) \( x + x^3 - 1 = 0 \)

This is a cubic equation. We can rewrite it as:

\[ x^3 + x - 1 = 0 \]

We use Ramanujan's method to estimate a root. Let's apply Newton-Raphson method for a more precise approximation. 

Newton-Raphson formula:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

Let \( f(x) = x^3 + x - 1 \) and \( f'(x) = 3x^2 + 1 \).

Start with an initial guess, \( x_0 = 0.5 \):

1. \( x_1 = 0.5 - \frac{0.5^3 + 0.5 - 1}{3 \cdot 0.5^2 + 1} = 0.5 - \frac{0.125 + 0.5 - 1}{0.75 + 1} = 0.5 - \frac{-0.375}{1.75} = 0.5 + 0.214 = 0.714 \)

2. \( x_2 = 0.714 - \frac{0.714^3 + 0.714 - 1}{3 \cdot 0.714^2 + 1} \approx 0.714 - \frac{0.364 + 0.714 - 1}{1.528 + 1} \approx 0.714 - \frac{0.078}{2.528} \approx 0.683 \)

3. Further iterations will refine the root, but the smallest root is approximately \( \boxed{0.683} \).

### (c) \( \sin x + x - 1 = 0 \)

This is a transcendental equation. We can apply Newton-Raphson method to approximate the root.

Let \( f(x) = \sin x + x - 1 \) and \( f'(x) = \cos x + 1 \).

Start with an initial guess, \( x_0 = 0.5 \):

1. \( x_1 = 0.5 - \frac{\sin(0.5) + 0.5 - 1}{\cos(0.5) + 1} \approx 0.5 - \frac{0.479 + 0.5 - 1}{0.877 + 1} = 0.5 - \frac{-0.021}{1.877} = 0.5 + 0.011 = 0.511 \)

2. \( x_2 = 0.511 - \frac{\sin(0.511) + 0.511 - 1}{\cos(0.511) + 1} \approx 0.511 - \frac{0.488 + 0.511 - 1}{0.858 + 1} = 0.511 - \frac{-0.001}{1.858} \approx 0.511 \)

The smallest root is approximately \( \boxed{0.511} \).

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If you want further details or have any questions, please let me know. Here are some related questions you might find interesting:

1. How to derive the Rational Root Theorem?
2. What are the steps of the Newton-Raphson method?
3. Can Ramanujan's method be applied to higher-degree polynomials?
4. How to solve transcendental equations using numerical methods?
5. What are other methods for solving cubic equations?

**Tip:** When dealing with transcendental equations, graphical methods can also be useful to estimate the roots before applying numerical methods like Newton-Raphson.

Explore how Ramanujan’s method is applied to find the smallest roots of cubic and transcendental equations. Includes detailed solutions for equations like x^3 - 6x^2 + 11x - 6 = 0, x + x^3 - 1 = 0, and sin x + x - 1 = 0 using Newton-Raphson method and the Rational Root Theorem.

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