Question

Solve:
----> (2-y)^3

Answer 1

it has to be equal to something to solve it

Answer 2

Its to the power of 3 -_-

Answer 3

yes... but that's not an equation... an equation needs an equal sign to be able to solve it.

do you need to simplify, distribute... substitute... or something else?

Answer 4

factorise

Answer 5

idk man its my HW lol

Answer 6

its already factorised.... a binomial of degree 3

Answer 7

ya true

Answer 8

in class we solved a question like this i just didnt understand

Answer 9

(2 - y)^3 = 27

is an equation you can solve

Answer 10

something like (2-y)(2-y)(2-y)

Answer 11

did you take any notes?

Answer 12

we all can solve (2-y)^2 right

Answer 13

just idk how with ^3

Answer 14

maybe you need to expand and write the polynomial?

Answer 15

(2-y)^3 = (2-y)^2*(2-y)
what is (2-y)^2 =

Answer 16

right okay we r getting somewhere

Answer 17

(2-y)^2 = (2-y)(2-y)
= 4-2y-2y-y^2

Answer 18

=4-4y-y^2

Answer 19

wat now :P

Answer 20

when you get that now multiply by (2-y)
remember the distributive property you just used now you have 3 terms

Answer 21

and how exactly ?

Answer 22

4-4y-y^2 (2-y) like so ?

Answer 23

(2-y)^2 = 4 -2y + y^2
(2-y)^3 = (2-y)(2-y)^2 = (2-y)(4 - 2y + y^2)
= 2(4 - 2y + y^2) -y(4 - 2y + y^2)
yes

Answer 24

advise to use ( ) otherwise think only -y^2 *(2-y)

Answer 25

continue from where I was when I answered yes

Answer 26

alright sure thanks

Answer 27

so like this (2-y)(4-2y+y^2)
8-4y+2y^2-4y+2y^2-y^3 ?

Answer 28

simplify by grouping like terms
common practice to rearrange in order of degree (highest to lowest etc)

are you back in class now

Answer 29

no at home

Answer 30

ok group like terms

Answer 31

ok i think i got the answere

Answer 32

8-8y+4y^2-Y^3 final answer ? how does it look?

Answer 33

I have the same i need to check

Answer 34

great, most likely correct then :D

Answer 35

8 -12y + 6 y^2 + y^3

Answer 36

yes found the error!!

Answer 37

final correct answer is
8 -12y + 6y^2 - y^3

Answer 38

@IDK69 the method is correct
my error was using (4-2y + y^2) instead of (4-4y + y^2)

Answer 39

I'm so sorry if I let you get it wrong. you expanded correctly. we had the same answer because you continued on my wrong entry. I knew something was wrong because of the pattern for the result.

Answer 40

ohh ok thats no problem appreciate the help