Prove this equation is one to one

[Painguy]

Homework Statement

$$y=(x+1)^3$$

Homework Equations

f(a)=f(b)
a=b

The Attempt at a Solution

$$(a+1)^3 =(b+1)^3$$
$$1+3 a+3a^2+a^3=1+3b+3b^2+b^3$$

This is where I get stuck.

 

[SammyS]

How about: if $(a+1)^3 =(b+1)^3\,,$ then $(a+1)^3-(b+1)^3=0\,.$

Then factor into difference of cubes.

 

[Dick]

Do you know any calculus? That's the easy way to prove this. Show y is an always increasing function of x.

 

[Painguy]

How about: if $(a+1)^3 =(b+1)^3\,,$ then $(a+1)^3-(b+1)^3=0\,.$

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 $$(x^3)+1$$ rather than $$(x+1)^3$$ I'm probably forgetting something here :P

Do you know any calculus? That's the easy way to prove this. Show y is an always increasing function of x.

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.

 

[SammyS]

The difference of cubes is:

p³ - r³ = (p-r)(p²+pr+r²)

If (p-r)(p²+pr+r²) = 0,

Then either p=r,

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

 

[Mark44]

m³ - n³ = (m - n)(m² + mn + n²)

m³ + n³ = (m + n)(m² - mn + n²)

 

[Deveno]

one could approach this another way:

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) = x³ is injective, and

g(x) = x+1 is injective.

 

[xaos]

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...