Question

HELP ME PLS:)
-(-2x^6y^7)^3 (x^2-y^2)^3

Answer 1

I'm here again! hahah.

Answer 2

how great GOD is, hehehe.. so lucky today..

Answer 3

\[(8x^{18}y^{21}) (x^4 -2x^2y^2 +y^4)(x^2 - y^2) \]

Answer 4

Hahah. I know, right?

Answer 5

LOL.

Answer 6

Then

Answer 7

then?:)

Answer 8

\[(8x^{18}y^{21})(x^6 -2x^4y^2 +x^2y^4 -x^4y^2 +2x^2y^4-y^6)\]

Answer 9

in \[(2x^6-y^7)^3\] i get

8x^18-12x^12y^7+6x^6y^14-y^21

i think im wrong

Answer 10

then just distribute it. I'm tired of typing. :))

Answer 11

how did you get that?

Answer 12

\[ 8x^18- 4x^12y^7 +4x^{12}y^7 -2x^6y^{14} + 2x^6y^14-y^21 \]

Answer 13

my ans. in (x^2-y^2)^3
is
\[x^6-3x^4y^2+3x^2y^4-y^6 \]

Answer 14

I think. LOL

Answer 15

Oh right right. LOL. Sorry.

Answer 16

am i right?

Answer 17

Yeah. And I'm wrong.
XD

Answer 18

what is your goal? multiply out?

Answer 19

I gotta go now. For real. hahah.

Answer 20

satellite, yeah!

Answer 21

ok first off first term is
\[8x^{18}y^{21}\]

Answer 22

then?

Answer 23

that is just cubing the first term. then you have to expand the second term. then you have to multiply

Answer 24

i cant clearly understand:(

Answer 25

ok you have two things here. one is
\[-(-2x^6y^7)^3\] yes?

Answer 26

and when we cube it we get what i wrote above namely
\[8x^{18}y^{21}\]

Answer 27

multiply the exponents and cube the 2 and there are two minus signs so we remove them

Answer 28

the next term is
\[(x^2-y^2)^3\] and when you cube this thing you get
\[x^6-3 x^4 y^2+3 x^2 y^4-y^6\]

Answer 29

you can multiply twice if you like or you can use
\[(a+b)^3=a^3+3a^2b+3ab^3+b^3\] with
\[a=x^2,b=-y^2\]

Answer 30

how are we doing so far?

Answer 31

it doesnt mean that im going to multiply the terms in itself? for example 1st term... am i not going to multiply by itself thrice? when it doesnt have operations?

Answer 32

heheheh:) what's the next step?

Answer 33

not sure exactly what you mean but
\[(x^2-y^2)^3=(x^2-y^2)(x^2-y^2)(x^2-y^2)\] and when you multiply out and combine like terms you get
\[x^6-3 x^4 y^2+3 x^2 y^4-y^6\]

Answer 34

last step is to write
\[8x^{18}y^{21}(x^6-3 x^4 y^2+3 x^2 y^4-y^6)\] and multiply out using distributive law

Answer 35

i get that.. now I know.. thank you:) i already forgot my 2nd yr. lesson.. thank you very much:)

Answer 36

just do the for multiplications adding the exponents as you go. when you are finished you should have
\[8 x^{24} y^{21}-24 x^{22} y^{23}+24 x^{20} y^{25}-8 x^{18} y^{27}\]

Answer 37

good luck

Answer 38

is that when we are adding in algebraic ex. we cannot add the terms if they have different exponents?

Answer 39

that is the final answer?

Answer 40

this is the final answer

Answer 41

yes you cannot add the terms if the have different exponents....the terms are linearly independent