Question

I have to show that y=x and so far I've simplified it to this: __(in comments)__. Please help, I don't know where to go from here. Thanks :)

Answer 1

\[x ^{3}+xy+y ^{2}=-1\]

Answer 2

what exactly are you trying to do here ?

Answer 3

show that x=y

Answer 4

you wrote "so far I've simplified it to this" so I'm assuming there's an initial problem that you didn't post? or is \(x ^{3}+xy+y ^{2}=-1\) the original problem?

Answer 5

oh...sorry, we have to show (from the definition) that
\[f(x) = x ^{3} + x \]
is a 1-1 function..
so I let
\[x ^{3}+x=y^2+y\]
then said...
\[x ^{3}-y ^{3}=y-x\]
Then...
\[(x-y)(x ^{2}+xy+y ^{2}) = y-x\]
then...
\[x ^{3}+xy+y ^{2}=\frac{ -(x-y) }{ (x-y) }\]

Answer 6

one of your powers is off i think

Answer 7

argh! sorry...I just realised that...
it should be x^2+xy+y^2=-1

Answer 8

\[\large (x-y)(x ^{2}+xy+y ^{2}) = y-x \iff x^{\color{Red}{2}}+xy+y^2 = -1\]

Answer 9

when \(x\ne y\)

Answer 10

otherwise you won't be able to divide x-y and cancel it

Answer 11

oh..well if I can't cancel then how do I go about it.

Answer 12

Notice that x^2+xy+y^2 = (x+y)^2 - xy

familiar with AM-GM inequality ?

Answer 13

no...sorry

Answer 14

Okay, what we're trying basically is :

\[\large (x-y)(x ^{2}+xy+y ^{2}) = y-x \\ \large (x-y)( x^{\color{Red}{2}}+xy+y^2 +1)= 0 \]

Answer 15

By zero product property, either the first factor or the second factor or both can equal 0. o if you prove that the second factor can never equal 0, you're done.

Answer 16

wait...how did you get that second line...to equal zero. I don't see what you did.

Answer 17

factored `x-y`

Answer 18

\[\large (x-y)(x ^{2}+xy+y ^{2}) = y-x \\ \large (x-y)(x ^{2}+xy+y ^{2})+(x-y) =0 \\ \large (x-y)( x^{\color{Red}{2}}+xy+y^2 +1)= 0\]

Answer 19

oh thanks! so.. (x-y) and (x^2+xy+y^2 +1) both have to equal zero for x=y right?

Answer 20

Not exactly

Answer 21

tell me this :

2*3 = 0

is that be true ?

Answer 22

no

Answer 23

what about this :

a*3 = 0

can this be true ever ?

Answer 24

oh so is either (x-y) or ('the other factor') equal to 0.

Answer 25

you might be seeing it already

a*b = 0

means,
either a or b has to be 0

Answer 26

exactly!

Answer 27

awesome! Thanks so much for your help :)

Answer 28

if we could show that "the other factor" is never 0, then that forces the first factor to equal 0

Answer 29

can you show that "the other factor" is never 0 ?

Answer 30

it is easy to show using AM-GM inequality, but since you don't want to use it, you need to find some other means

Answer 31

umm...well
\[(x+y) = + or - \sqrt{-1-xy}\]
does that show anything?

Answer 32

Here is how I would work it
\[\large (x-y)(x ^{2}+xy+y ^{2}) = y-x \\ \large (x-y)(x ^{2}+xy+y ^{2})+(x-y) =0 \\ \large (x-y)( x^{\color{Red}{2}}+xy+y^2 +1)= 0 \\\large (x-y)=0 ~\lor ~( x^{\color{Red}{2}}+xy+y^2 +1)= 0\]
the minimum value of \(\large x^2 + xy+y^2+1 \) is \(\large 1\), so it can never equal \(\large 0\). That forces the first factor to equal \(0\) :
\[\large x-y = 0 \implies x = y\]

Answer 33

When you say "the minimum value of x^2 + xy+y^2+1 is 1", can't it go lower? x could be some very large negative number while y is positive and I can see it dipping below 1.

Answer 34

By AM-GM inequality \(\large (x+y)^2 \ge 4xy\)

Answer 35

oh i guess the x^2 counterbalances that

Answer 36

\[\large x^2 + xy+y^2 = (x+y)^2 - xy \ge 0\]

But the OP is not familiar with this inequality so I thought i better skip the explanation and let him figure out his own way of showing this :)

Answer 37

what I'd do is solve for x in x^2 + xy + y^2 + 1 = 0 by using the quadratic formula

x^2 + xy + y^2 + 1 = 0 or x^2 + y*x + y^2 + 1 = 0 is in the form ax^2 + bx + c = 0 where

a = 1
b = y
c = y^2 + 1

Answer 38

actually, you can focus on the discriminant formula

D = b^2 - 4ac

if D < 0, then there won't be any real-number solutions to the quadratic

Answer 39

that works, its just a quadratic in two variables after all

Answer 40

oh ok cool!