Evaluate the following limits: x 1 (7x2 - 4x - 3)/(3x2 - 4x + 1) and x -2 (x4 + 5x3 + 6x2)/(x2(x+1) - 4(x+1))

$$\lim_{x \to 1} \frac{7x^2 - 4x - 3}{3x^2 - 4x + 1}$$

Substituting $x=1$ directly gives $\frac{7(1)^2 - 4(1) - 3}{3(1)^2 - 4(1) + 1} = \frac{7-4-3}{3-4+1} = \frac{0}{0}$, which is an indeterminate form. We factor the numerator and the denominator.
Since $x=1$ is a root for both, $(x-1)$ is a factor.
Factor the numerator: $7x^2 - 4x - 3 = (x-1)(7x+3)$.
Factor the denominator: $3x^2 - 4x + 1 = (x-1)(3x-1)$.
Now, substitute these factored forms into the limit expression:

$$\lim_{x \to 1} \frac{(x-1)(7x+3)}{(x-1)(3x-1)}$$

Cancel out the common factor $(x-1)$ (since $x \neq 1$ as $x \to 1$):

$$\lim_{x \to 1} \frac{7x+3}{3x-1}$$

Now, substitute $x=1$:

$$\frac{7(1)+3}{3(1)-1} = \frac{7+3}{3-1} = \frac{10}{2} = 5$$

The value of the first limit is $\boxed{5}$.

$$\lim_{x \to -2} \frac{x^4 + 5x^3 + 6x^2}{x^2(x+1) - 4(x+1)}$$

Substituting $x=-2$ directly gives $\frac{(-2)^4 + 5(-2)^3 + 6(-2)^2}{(-2)^2(-2+1) - 4(-2+1)} = \frac{16 - 40 + 24}{4(-1) - 4(-1)} = \frac{0}{0}$, which is an indeterminate form. We need to simplify the expression.
Factor the numerator:
$x^4 + 5x^3 + 6x^2 = x^2(x^2 + 5x + 6) = x^2(x+2)(x+3)$.
Factor the denominator:
$x^2(x+1) - 4(x+1) = (x+1)(x^2-4) = (x+1)(x-2)(x+2)$.
Now, substitute these factored forms into the limit expression:

$$\lim_{x \to -2} \frac{x^2(x+2)(x+3)}{(x+1)(x-2)(x+2)}$$

Cancel out the common factor $(x+2)$ (since $x \neq -2$ as $x \to -2$):

$$\lim_{x \to -2} \frac{x^2(x+3)}{(x+1)(x-2)}$$

Now, substitute $x=-2$:

$$\frac{(-2)^2(-2+3)}{(-2+1)(-2-2)} = \frac{4(1)}{(-1)(-4)} = \frac{4}{4} = 1$$

The value of the second limit is $\boxed{1}$.

$$\lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{\sqrt{x} - 1}$$

Substituting $x=1$ directly gives $\frac{\sqrt[3]{1} - 1}{\sqrt{1} - 1} = \frac{1-1}{1-1} = \frac{0}{0}$, which is an indeterminate form.
To simplify, let $x = t^6$. As $x \to 1$, $t \to 1$.
Then $\sqrt[3]{x} = \sqrt[3]{t^6} = t^2$ and $\sqrt{x} = \sqrt{t^6} = t^3$.
Substitute these into the limit expression:

$$\lim_{t \to 1} \frac{t^2 - 1}{t^3 - 1}$$

Factor the numerator using the difference of squares formula $a^2-b^2=(a-b)(a+b)$:
$t^2 - 1 = (t-1)(t+1)$.
Factor the denominator using the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2)$:
$t^3 - 1 = (t-1)(t^2+t+1)$.
Now, substitute these factored forms into the limit expression:

$$\lim_{t \to 1} \frac{(t-1)(t+1)}{(t-1)(t^2+t+1)}$$

Cancel out the common factor $(t-1)$ (since $t \neq 1$ as $t \to 1$):

$$\lim_{t \to 1} \frac{t+1}{t^2+t+1}$$

Now, substitute $t=1$:

$$\frac{1+1}{1^2+1+1} = \frac{2}{1+1+1} = \frac{2}{3}$$

The value of the third limit is $\boxed{\frac{2}{3}}$.