Question

kindly solve this in special products:
(1/2r^2-1/4r^2-r^3)^2

Answer 1

\[(\frac{1}{2r^2}-\frac{1}{4r^2}-r^3)^2\] is this what you want?

Answer 2

because if so before you square you can combine the first two terms as
\[(\frac{1}{2r^2}-\frac{1}{4r^2}-r^3)^2=(\frac{1}{4r^2}-r^3)^2\]

Answer 3

\[(1/2r^3-1/4r^2-r^3)^2\]

Answer 4

thats the right given.. hehehe

Answer 5

oooooooooooooooh

Answer 6

sorry, .my bad:(

Answer 7

that's ok but what are you supposed to do with this? multiply it out?

Answer 8

yes, but using special products..

Answer 9

i don't know what that means in this case. you have
\[(a+b+c)^2=a^2+2 a b+2 a c+b^2+2 b c+c^2\] is this the "special product" you are supposed to use?

Answer 10

in whatever order you choose. so if you do it here i think the substitution will give
\[r^6+\frac{1}{4 r^6}-\frac{1}{4 r^5}+\frac{1}{16 r^4}+\frac{r}{2}-1\]

Answer 11

how did you get that?