solve (x-2y)^3 - (x-2y)^5?

Question

its MINUS not divide or multipy (which would be EASY)

my teacher gave me a hint - think of it as (p)^3 - (p)^3

simplify this,
i dont want an answer of what x or y is equal to, just simplify

and no (p)^3 - (p)^5 does not equal 0 ....

Answer 1

yep your teachers hint was p^3 - p^5

you take the common factor out to get

p^3 (1- p^2)

to put the innards of your original equation in at this point gives

(x-2y)^3 [1- (x-2y)^2]

= (x-2y)^3 [(1-x+2y)(1+x-2y)]

you can give this as your answer but also give it multiplied out so that your teacher can see you have done at least some of it yourself

Answer 2

ok
(x-2y)^3 - (x-2y)^5=(x-2y)^3 [1-(x-2y)^2]
now
what do you want to solve? do you want to solve:
(x-2y)^3 - (x-2y)^5=0????

if p=x-2y
then
we have
p^3-p^5=p^3(1-p^2)=0
means that p=0
or 1-p^2=0

then
p=x-2y=0
or
p^2=1, p=1 or p=-1, which means that
x-2y=1 or x-2y=-1

Answer 3

(x - 2y)^3 - (x - 2y)^5 =

(x - 2y)^3 * [1 - (x - 2y)^2] =

(x - 2y)^3 * ( 1 - x + 2y) * (1 + x - 2y)

Answer 4

x^-2 - y^-2

Answer 5

a3 - b3 = (a - b)( a² + ab +b²)