Prove that, for all real numbers x and y, if x < y, then x^3 < y^3. It does not suffice to merely prove this for positive x and y; your proof must handle all cases.

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- Bee_see

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- jim_thompson5910

x^3 < y^3
x^3 - y^3 < 0
(x-y)*(x^2 + xy + y^2) < 0 ... difference of cubes factoring rule
if x < y, then x-y < 0. So x-y is negative while x^2 + xy + y^2 is always positive
so if x < y, then (x-y)*(x^2 + xy + y^2) < 0 is always true

- Bee_see

how do you know x-y is negative and x^2+xy+y^2 is always positive?

- jim_thompson5910

x < y leads to x-y < 0 which means x-y is negative

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## More answers

- jim_thompson5910

x^2 is positive
y^2 is positive
since x < y, this means |x*y| < y^2
so even if x*y was negative, the positive x^2 and y^2 terms counterbalance and make x^2+xy+y^2 always positive

- Bee_see

but if I were to say x=5 and y=4, then don't I get positive 1?

- jim_thompson5910

keep the inequality x < y in mind

- Bee_see

so like, 2 < 3 2-3=-1?

- jim_thompson5910

yeah that's one example

- Bee_see

how does it fulfill all the cases?

- jim_thompson5910

you use an algebraic proof like I did above to show it works in every case (as long as x < y)

- Bee_see

how did you know that you wanted the equation to equal 0? (in the beginning)

- jim_thompson5910

you mean in my step 2? I subtracted y^3 from both sides to make the right side 0

- jim_thompson5910

btw it's not an equation. It's an inequality

- Bee_see

I mean this step: x^3 - y^3 < 0

- jim_thompson5910

I subtracted y^3 from both sides of x^3 < y^3

- jim_thompson5910

I guess my proof in my first post is me thinking backwards

- jim_thompson5910

but it's still valid in a way

- Bee_see

No, I mean...so I guess you are not supposed to go from x < y to x^3 < y^3?

- jim_thompson5910

Here's an attempt at doing the proof in the right order (notice how it's in reverse order of my first proof)
if x < y, then x-y < 0
multiply (x-y) with (x^2+xy+y^2). This result is always negative because x-y < 0 and x^2+xy+y^2 > 0
(x-y)*(x^2+xy+y^2) < 0
x*(x^2+xy+y^2)-y*(x^2+xy+y^2) < 0
x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 < 0
x^3 - y^3 < 0
x^3 < y^3

- Bee_see

OK. I see.

- Bee_see

Thanks for the help.

- jim_thompson5910

no problem

- jim_thompson5910

here's a page showing why x^2 + xy + y^2 is always positive (assuming x and y are nonzero)
http://math.stackexchange.com/questions/349940/if-x-neq-0-y-neq-0-then-x2xyy2-is

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