This is from implicit differentiation and I'm curious as to why something like
$\dfrac{y^3-y^2\times\left(\dfrac{-1}{y^3}\right)}{y^6}$
Equals
$\dfrac{y^4 + 1}{y^7}$
I understand why the denominator is y to the seventh but unsure why the first $y^3$ becomes $y^4$ because we divide only the second $y^2$ by $y^3$
Is it because we times the numerator and denominator by $y$?
I got this out of a book and it doesn't really explain the answer
We have $$ \frac{y^3 - y^2 \cdot \frac{-1}{y^3}}{y^6} = \frac{y^3 + \frac{1}{y}}{y^6} = \frac{y}{y} \cdot \frac{y^3 + \frac{1}{y}}{y^6} = \frac{y^4 + 1}{y^7}. $$ This sort of manipulation is frequently done when one wants to "simplify fractions", where "simple" means that the numerator and denominator shouldn't themselves be fractions.