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I'm currently going through Spivak's Calculus, and I got across the following problem:

Prove that $x^3+y^3=(x+y)(x^2-xy+y^2)$.

To prove it, the author wants to use another problem that was proved earlier, i.e $x^3-y^3 = (x-y)(x^2+xy+y^2)$

The solution says to simply replace $y$ with $-y$ in the above equation. But $y$ can't be equal to $-y$ as long as $y \neq 0$, so how is this a legal operation?

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    $\begingroup$ For "=/=" use "dollar backslash neq dollar". $\endgroup$ – barak manos Jan 4 '16 at 8:02
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    $\begingroup$ Go back to the proof of (typo) $x^3-y^3=(x-y)(x^2+xy+y^2)$. Use almost word for word the same argument to show $u^3-v^3=(u-v)(u^2+uv+v^2)$. Now replace $u$ by $x$ and $v$ by $-y$. $\endgroup$ – André Nicolas Jan 4 '16 at 8:04
  • $\begingroup$ y=-y is a little bit confusing, try to substitute y = -u and you will see. $\endgroup$ – ZTransformer Jan 4 '16 at 8:04
  • $\begingroup$ since the formula for $x^3 - y^3$ holds for all real numbers, it may prudent to think about $x^3 + y^3$ as $x^3 - (-y)^3$ where y is a positive real number (it is obviously negated by the $-$ sign). Then replace $y$ with $-y$ on the RHS and see how it works out. (Think about it like inputting variables in a function, that is, we input $-y$ instead of $y$) $\endgroup$ – Airdish Jan 4 '16 at 17:19
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The author is saying: $y$ is not a particular number, that equation holds for every real number. So it works for $y=3$ but also for $y = -3$ or $y = \pi$ or whatever.

So you can "substitute" $-y$ instead of $y$ because, even if the two numbers are different, the relation still holds

Another way of doing it would be: call $y = -z$. Look at the result you get.

Clearly now you'll find a relation in $z$ and not in $y$. But since $z$ and $y$ are only names, you can always call $z = y$ to get the "correct" result. What we did was essentially substituting $y = -y$ but maybe now it's a little bit more clear

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Ant's answer is correct, but here's another way of saying the same thing, with a telling example.

Spivak's operation is in fact "legal," as you say, because it is just an instance of what a logician would call the "law of substitution."

As another example, presumably you know how a difference of squares factors:

$$x^2-y^2=(x-y)(x+y)$$

But this identity holds for all real numbers $x$ and $y$; $x$ and $y$ can stand for whatever real number you want. So we should really think of this equation as shorthand for infinitely many different factorizations, each given by substituting a different value for $x$ and $y$. For example, substituting $x=a+b$ and $y=c+d+e$ gives the seemingly more difficult result

$$(a+b)^2-(c+d+e)^2=(a+b-c-d-e)(a+b+c+d+e)$$

But really this equation follows from the first by the substitution principle. That's exactly the sort of argument Spivak is giving here.

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What substituting $y=-y$ essentially means as follows:

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

Hence $x^3 + y^3$ = $(x - (-y))(x^2 + x(-y) + y^2)$

= $(x+y)(x^2 -xy +x^2)$

Well it's as simple as that. Basically you just input any arbitrary negative number and you get this result. Hope this helped explain the intuition behind the substitution.

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