Proof for x^n-y^n=(x-y)(x^n-1+ +y^n-1)

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In summary, the proof for x^n-y^n=(x-y)(x^n-1+...+y^n-1) involves using induction. Starting with the base case of n=1, the equation is proven to be true. Then, assuming it is true for n=k, the equation is expanded to n=k+1 and rearranged to match the desired form. By noticing a pattern and using algebraic manipulation, the equation is proven to be true for all n\geq1.
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moviefan91
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Proof for x^n-y^n=(x-y)(x^n-1+...+y^n-1)

Homework Statement


The question asks to prove that for any n[itex]\geq[/itex]1,
[itex]x^{n}[/itex]-[itex]y^{n}[/itex]=(x-y)([itex]x^{n-1}[/itex]+[itex]x^{n-2}[/itex]y+...+[itex]y^{n-1}[/itex])


Homework Equations


[itex]x^{n}[/itex]-[itex]y^{n}[/itex]=(x-y)([itex]x^{n-1}[/itex]+[itex]x^{n-2}[/itex]y+...+[itex]y^{n-1}[/itex])


The Attempt at a Solution



So far, I used induction.
So for n=1, x-y=x-y

Second step, I assume that n=k is true:
[itex]x^{k}[/itex]-[itex]y^{k}[/itex]=(x-y)([itex]x^{k-1}[/itex]+[itex]x^{k-2}[/itex]y+...+[itex]y^{k-1}[/itex])

I get stuck at n=k+1.

[itex]x^{k+1}[/itex]-[itex]y^{k+1}[/itex]=(x-y)([itex]x^{k}[/itex]+[itex]x^{k-1}[/itex]y+...+[itex]y^{k}[/itex])

When I expand RHS, I get:
[itex]x^{k+1}[/itex]-[itex]x^{k}[/itex]y+[itex]x^k{}[/itex]y-[itex]x^{k-1}[/itex][itex]y^{2}[/itex]+...+x[itex]y^{k}[/itex]-[itex]y^{k+1}[/itex]

I think that I need to cancel things so I can be left only with [itex]x^{k+1}[/itex]-[itex]y^{k+1}[/itex], but I always have terms in the middle which do not cancel out.

What am I doing wrong?
 
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you need to work from the the n case to the n+1 (or vice versa)

i haven't worked it, but how about noticing:
[tex](x+y)(x^{k} -y^{k}) = x^{k+1} -xy^{k} -x^{k}y -y^{k+1}[/tex]

then you have
[tex]x^{k+1} -y^{k+1} = (x+y)(x^{k} -y^{k})-x^{k}y +xy^{k}[/tex]

then see if you can work it into the required form...
 

Related to Proof for x^n-y^n=(x-y)(x^n-1+ +y^n-1)

1. What is the formula for x^n-y^n?

The formula for x^n-y^n is (x-y)(x^n-1+x^n-2y+...+y^n-1).

2. How is the formula for x^n-y^n derived?

The formula for x^n-y^n is derived from the binomial theorem, which states that (a+b)^n = a^n + na^(n-1)b + n(n-1)/2!a^(n-2)b^2 + ... + b^n. By substituting a with x and b with -y, we get (x-y)^n = x^n + nx^(n-1)(-y) + n(n-1)/2!x^(n-2)(-y)^2 + ... + (-y)^n. Simplifying this equation gives us the formula for x^n-y^n.

3. What is the significance of x^n-y^n in mathematics?

The formula for x^n-y^n has various applications in mathematics, including in algebraic equations, number theory, and geometry. It allows us to factorize expressions and solve for unknown variables, making it a fundamental tool in many mathematical problems.

4. Can the formula for x^n-y^n be applied to any value of n?

Yes, the formula for x^n-y^n can be applied to any value of n, whether it is a positive integer, negative integer, or fraction. However, for non-integer values of n, the formula may not yield a whole number as a result.

5. How does the formula for x^n-y^n relate to the concept of difference of squares?

The formula for x^n-y^n is a generalization of the concept of difference of squares, which states that a^2-b^2 = (a+b)(a-b). When n=2, the formula for x^n-y^n simplifies to (x-y)(x+y), which is equivalent to the difference of squares formula. Therefore, we can say that the formula for x^n-y^n is an extension of the difference of squares concept.

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