Hi Cockchestner007,

\(\displaystyle \frac{y^3-1}{y^2+y-6} \div \frac{y-1}{3+y}\)

Invert the divisor and multiply.

\(\displaystyle \frac{y^3-1}{y^2+y-6} \cdot \frac{y+3}{y-1}\)

\(\displaystyle \frac{(y-1)(y^2+y+1)}{(y+3)(y-2)} \cdot \frac{y+3}{y-1}\)

The "restrictions of the variable" means "what would y have to be to make the denominator of either fraction zero?"

Those restrictions would be: **y cannot be -3, 2, or 1**

Simplifying we arrive at:

\(\displaystyle \frac{y^2+y+1}{y-2}\)

The only restriction after simplification is **y cannot be 2**.

If we were talking about rational functions, this would be a vertical asymptote.

The other two restrictions would be points of discontinuity on the graph.

But, this may be too much information since these are only expressions.