How do you find the product of (12c^3)/(21b)*(14b^2)/(6c)?

Jun 15, 2018

See a solution process below:

Explanation:

First, factor the expression as:

$\frac{6 \cdot 2 \cdot c \cdot {c}^{2}}{3 \cdot 7 \cdot b} \cdot \frac{2 \cdot 7 \cdot b \cdot b}{6 \cdot c}$

Next, cancel common terms in the numerator and denominator:

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} \cdot 2 \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{c}}} \cdot {c}^{2}}{3 \cdot \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{7}}} \cdot \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{b}}}} \cdot \frac{2 \cdot \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{7}}} \cdot \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{b}}} \cdot b}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{c}}}} \implies$

$\frac{2 \cdot {c}^{2}}{3} \cdot \frac{2 \cdot b}{1} \implies$

$\frac{2 \cdot {c}^{2} \cdot 2 \cdot b}{3 \cdot 1} \implies$

$\frac{4 b {c}^{2}}{3}$