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I can't seem to get my head around the underlined in the proof attached being correct. Plug x as 3, y as 4 and n as 2 and you end up with -7=-14 for example.

Proof

What is going on?

Thank you! :)

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    Where do you get the $14$ from? $(3-4)(3^1\cdot 4^0 + 3^0\cdot 4^1) = (-1)\cdot 7 = -7$. – 2017-01-16
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    Welcome to math stack exchange! – 2017-01-16
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    Just out of curiosity: "In view of the above theorem"... what is this theorem? – 2017-01-16
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    $$3^2-4^2=(3-4)(3+4)\iff-16=-1\cdot7\iff-7=-7$$ – 2017-01-16
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    @sranthrop continuity of addition, no doubt – 2017-01-16
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    @Omnomnomnom: I think so too, but sometimes they prove that multiplication is also continuous and then the polynomial-example follows exactly like here, even if it is then not necessary. That's what I wanted to point out just as a little remark. – 2017-01-16
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    @sranthrop Just that you can add, multiply and divide continous functions to get another continuous function. – 2017-01-16
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    Ok, so then this proof is - as I suspected - not necessary, since $x^n=x\cdot\ldots\cdot x$ is the product of the continuous function $x\mapsto x$ with itself, $n$-times. – 2017-01-16

3 Answers 3

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Putting $x=3$, $y=4$ and $n=2$ gives the same answer: $$3^2-4^2=-7$$ $$(3-2)(3+4)=-7$$

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    Well this is embarassing. Ha, thanks! – 2017-01-16
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As your left hand side is correct.

Right hand side -

$(3 - 4)(3^1 + 3^0 \cdot 4^1)$

= $(-1)(3 + 1 \cdot 4)$

= (-1)(3 + 4)

= -7

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    Mine pleasure :-) – 2017-01-16
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$$ \begin{gathered} x^{\,n} - y^{\,n} \quad \left| \begin{gathered} \;y \ne x \hfill \\ \;x \ne 0 \hfill \\ \end{gathered} \right.\quad = x^{\,n} \left( {1 - \left( {\frac{y} {x}} \right)^{\,n} } \right) = x^{\,n} \left( {1 - \left( {\frac{y} {x}} \right)} \right)\left( {\frac{{1 - \left( {\frac{y} {x}} \right)^{\,n} }} {{1 - \left( {\frac{y} {x}} \right)}}} \right) \hfill \\ = x^{\,n} \left( {1 - \left( {\frac{y} {x}} \right)} \right)\left( {1 + \left( {\frac{y} {x}} \right) + \left( {\frac{y} {x}} \right)^2 + \cdots + \left( {\frac{y} {x}} \right)^{n - 1} } \right) = \hfill \\ = x\left( {1 - \left( {\frac{y} {x}} \right)} \right)x^{\,n - 1} \left( {1 + \left( {\frac{y} {x}} \right) + \left( {\frac{y} {x}} \right)^2 + \cdots + \left( {\frac{y} {x}} \right)^{n - 1} } \right) = \hfill \\ = \left( {x - y} \right)\left( {x^{\,n - 1} y^{\,0} + x^{\,n - 2} y^{\,1} + \cdots + x^{\,0} y^{\,n - 1} } \right) \hfill \\ \end{gathered} $$