- #1

Painguy

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## Homework Statement

[tex]y=(x+1)^3[/tex]

## Homework Equations

f(a)=f(b)

a=b

## The Attempt at a Solution

[tex](a+1)^3 =(b+1)^3[/tex]

[tex]1+3 a+3a^2+a^3=1+3b+3b^2+b^3[/tex]

This is where I get stuck.

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- Thread starter Painguy
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- #1

Painguy

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[tex]y=(x+1)^3[/tex]

f(a)=f(b)

a=b

[tex](a+1)^3 =(b+1)^3[/tex]

[tex]1+3 a+3a^2+a^3=1+3b+3b^2+b^3[/tex]

This is where I get stuck.

- #2

SammyS

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Then factor into difference of cubes.

- #3

Dick

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- #4

Painguy

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Then factor into difference of cubes.

I know what the difference of cubes is but I can't seem to figure out how to apply that to this case. I was under the impression that if I was to apply the difference of cubes that the equation would gabber to be in, the form [tex](x^3)+1[/tex] rather than [tex](x+1)^3[/tex] I'm probably forgetting something here :P

I haven't taken calculus yet but would it suggestion look like this? f(x)=(x+1)^3 I've already graphed it & seen that it's 1:1 but I'm looking to do this algebraically. Excuse me if I misinterpreted your post.

- #5

SammyS

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The difference of cubes is:

p^{3} - r^{3} = (p-r)(p^{2}+pr+r^{2})

If (p-r)(p^{2}+pr+r^{2}) = 0,

Then either p=r,

or [itex]\displaystyle p=\frac{-r\pm\sqrt{r^2-4r^2}}{2}[/itex], which is only a real number if r = 0.

p

If (p-r)(p

Then either p=r,

or [itex]\displaystyle p=\frac{-r\pm\sqrt{r^2-4r^2}}{2}[/itex], which is only a real number if r = 0.

Last edited:

- #6

Mark44

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m^{3} - n^{3} = (m - n)(m^{2} + mn + n^{2})

m^{3} + n^{3} = (m + n)(m^{2} - mn + n^{2})

m

- #7

Deveno

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if f:A→A is injective, and g:A→A is injective, then fog:A→A is injective.

proof:

suppose fog(x) = fog(y), that is:

f(g(x)) = f(g(y)).

since f is injective, g(x) = g(y).

since g is injective, x = y, QED.

now prove f(x) = x

g(x) = x+1 is injective.

- #8

xaos

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another path, i think the easiest mechanically but the least intuitive if you're given another definition, is to use the logical contrapositive equivalent of 1-to-1 (perhaps better known as proof by contradiction):

"x != y implies g o h(x) != g o h(y)".

here you don't need to show 1-to-1 of anything, only that the functions g and h are defined at x and y as you construct the conclusion f= g o h.

[edit] of course that ends up needing to show that g and h are indeed functions...

"x != y implies g o h(x) != g o h(y)".

here you don't need to show 1-to-1 of anything, only that the functions g and h are defined at x and y as you construct the conclusion f= g o h.

[edit] of course that ends up needing to show that g and h are indeed functions...

Last edited:

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