Simplify the given expression: $\frac{t^3 - 1}{7t^2 + 7t + 7}$.

1. Problem Description

Simplify the given expression: $\frac{t^3 - 1}{7t^2 + 7t + 7}$.

2. Solution Steps

First, factor the numerator, $t^3 - 1$, using the difference of cubes formula:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

In this case, $a = t$ and $b = 1$, so

$t^3 - 1 = (t - 1)(t^2 + t + 1)$

Next, factor the denominator, $7t^2 + 7t + 7$, by factoring out the common factor 7:

$7t^2 + 7t + 7 = 7(t^2 + t + 1)$

Now, we can rewrite the original expression as:

$\frac{t^3 - 1}{7t^2 + 7t + 7} = \frac{(t - 1)(t^2 + t + 1)}{7(t^2 + t + 1)}$

Since $t^2 + t + 1$ is a common factor in both the numerator and denominator, we can cancel it out, assuming $t^2 + t + 1 \neq 0$:

$\frac{(t - 1)(t^2 + t + 1)}{7(t^2 + t + 1)} = \frac{t - 1}{7}$

3. Final Answer

$\frac{t-1}{7}$