Let's solve the equations using Newton's method for both #25 and #27.

Problem #25:

$x^2 \sqrt{2} - x - x^2 = 1$ This can be simplified to: $x^2(\sqrt{2} - 1) - x - 1 = 0$

Newton's Method

Newton's method formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where $f(x) = x^2(\sqrt{2} - 1) - x - 1$.

1. Find $f(x)$: $f(x) = x^2(\sqrt{2} - 1) - x - 1$

2. Find $f'(x)$: $f'(x) = 2x(\sqrt{2} - 1) - 1$

We'll choose an initial guess $x_0$ by approximating the root graphically and apply Newton's method iteratively.

Problem #27:

$4e^{-x} \sin x = x^2 - x + 1$

Newton's Method

Here, let $f(x) = 4e^{-x} \sin x - (x^2 - x + 1)$.

1. Find $f(x)$: $f(x) = 4e^{-x} \sin x - (x^2 - x + 1)$

2. Find $f'(x)$: $f'(x) = 4e^{-x}(\cos x - \sin x) - (2x - 1)$

We'll also choose an initial guess $x_0$ for this equation and apply Newton's method iteratively.

Would you like me to proceed with the calculations, providing initial guesses and iterations for both equations?