6x - 3 / 7 + (x^3 - 1) when x = 1

1 answer

To evaluate the expression \( \frac{6x - 3}{7} + (x^3 - 1) \) when \( x = 1 \), we will substitute \( x = 1 \) into the expression.

First, let's substitute \( x = 1 \) into \( \frac{6x - 3}{7} \):

\[ \frac{6(1) - 3}{7} = \frac{6 - 3}{7} = \frac{3}{7} \]

Next, substitute \( x = 1 \) into \( (x^3 - 1) \):

\[ (1^3 - 1) = 1 - 1 = 0 \]

Now, we combine both parts:

\[ \frac{3}{7} + 0 = \frac{3}{7} \]

Thus, the evaluated expression when \( x = 1 \) is

\[ \boxed{\frac{3}{7}} \]