Simplify (49-y^2)/(6y^2+42y)*(48y^3)/(y^2-10y+21)

Rewrite ${\displaystyle 49}$ as ${\displaystyle 7^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=7}$ and ${\displaystyle b=y}$.

Factor ${\displaystyle 6y}$ out of ${\displaystyle 6y^2}$.

Factor ${\displaystyle 6y}$ out of ${\displaystyle 42y}$.

Factor ${\displaystyle 6y}$ out of ${\displaystyle 6y\left(y\right)+6y\left(7\right)}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 21}$ and whose sum is ${\displaystyle -10}$.

Write the factored form using these integers.

Combine.

Cancel the common factor of ${\displaystyle 48}$ and ${\displaystyle 6}$.

Cancel the common factor of ${\displaystyle y^3}$ and ${\displaystyle y}$.

Cancel the common factor of ${\displaystyle 7+y}$ and ${\displaystyle y+7}$.

Cancel the common factor of ${\displaystyle 7-y}$ and ${\displaystyle y-7}$.

Simplify the expression.