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

Question

Ethan Adkins · Accepted Answer

See a solution process below:

First, factor the expression as: #(6 * 2 * c * c^2)/(3 * 7 * b) * (2 * 7 * b * b)/(6 * c)# Next, cancel common terms in the numerator and denominator: #(color(red)(cancel(color(black)(6))) * 2 * color(blue)(cancel(color(black)(c))) * c^2)/(3 * color(green)(cancel(color(black)(7))) * color(purple)(cancel(color(black)(b)))) * (2 * color(green)(cancel(color(black)(7))) * color(purple)(cancel(color(black)(b))) * b)/(color(red)(cancel(color(black)(6))) * color(blue)(cancel(color(black)(c)))) =># #(2 * c^2)/3 * (2 * b)/1 =># #(2 * c^2 * 2 * b)/(3 * 1) =># #(4bc^2)/3#

Abigail Allen · Answer

undefined To find the product of (12c^3)/(21b)*(14b^2)/(6c), you can simplify the expression by canceling out common factors in the numerator and denominator.

First, cancel out the common factors between 12c^3 and 6c, which is 6c. This leaves us with (2c^2)/(21b)*(14b^2)/1.

Next, cancel out the common factors between 21b and 14b^2, which is 7b. This gives us (2c^2)/(3)*(2b)/1.

Finally, multiply the remaining terms together: (2c^2 * 2b)/(3 * 1) = (4bc^2)/3.

Therefore, the product of (12c^3)/(21b)*(14b^2)/(6c) simplifies to (4bc^2)/3.