Show that x^3=y^3 implies x=y

[ironman1478]

the question is from the book "elementary geometry from an advanced standpoint 3rd edition" by edwin e. moise

Homework Statement

Given x>0 and y>0, show that x^3 = y^3 => x = y. Does this hold for all every x and y?

Homework Equations

a^3-b^3=(a-b)(a^2+ab+b^2)=0

The Attempt at a Solution

so what i did was subtract y^3 from both sides to get
x^3-y^3 = 0

then i factored it out to
(x-y)(x^2+xy+y^2) = 0

because we know that x>0 and y>0, the second term (x^2+xy+y^2) is always positive. because of this (x-y) must equal zero
then we setup the equation x-y=0
x=y.

i think i did this correctly, but since i am teaching myself out of this book (i just want to learn more about geometry because i felt like i was never taught it well) i have no way of verifying if this is correct

 

[I like Serena]

Welcome to PF, ironman14781

Looks good!

Btw, this holds true for every x and y (real numbers).
Why would you think otherwise?

 

[ironman1478]

edit: nvm, i read the question wrong. i see what you and the question mean now lol

thanx

 

[I like Serena]

if y is negative and x is positive then their cubes can't be equal right?

No...?

But then the conditions do not hold either:
y>0
x^3=y^3
x=y

 

[HallsofIvy]

It is not a matter of "x< 0, y> 0". If x and y are any two numbers such that $x^3= y^3$, then x= y. It may be that x and y are both postive or that they are both negative (or both 0).
If x and y are both negative then xy is positive so it is still true that $x^2+ xy+ y^2$ is positive.