Algebra help please??

Question

how do you do ..

-9r ( -2r^2 ) ^3

the ^ = powers.

thanks!

Answer 1

note

power to a power is multiplication of power, so

r^2^3 = r^6

(-2)^3 = -8

so

-9r ( -2r^2 ) ^3 = -9r(-8r^6)

multiplication of same base is the addition of exponents so

r(r^6) = r^7

so

= 72r^7

Answer 2

First, since the ^3 applies only to (-2r^2) let's cube that

(r^2)^3 = r^6  (multiply the exponents).

(-2)^3 = -8

Now we have -9r (-8r^6)

minus times minus is plus so now we have

9r(8r^6) =

(9 x 8) x r x  r^6 = 72 x r x r^6  

(r = r^1, so r^1 x r^6 = r^7 (adding exponents)

72 r^7   <<<<<  final answer