WARNING:-
This question may have wrong tags.
HISTORY:-
After watching an old numberphile video(https://youtu.be/wymmCdLdPvM)(this video has nothing to do with my question) ,I got interested in the equation: $x^3+y^3+z^3=k$
Now being a high school student ,it is not likely that I am gonna prove some hundred years old uncracked problem surrounding the equation.
But you know, we all have false hopes, and so I started playing around with the beautiful equation.
After writing down the equation in a page, I remembered "Oo, I have to be smart to solve problems... ooooo." But I still kept playing with the equation and decided to write the sum of three cubes as a sum of 2 cubes
Here's what I did:
Assume that $x\neq -y$
$x^3+y^3+z^3=k$
$\implies x^3+y^3=k-z^3$
-eq(1)
$\implies (x+y)(x^2-xy+y^2)=k-z^3$
$\implies x^2-xy+y^2=\frac{k-z^3}{x+y}$
$\implies (x+y)^2=\frac{k-z^3}{x+y}+3xy$
$\implies (x+y)^3=k-z^3+3xy(x+y)$
$\implies (x+y)^3+z^3=k-3x^2y-3xy^2$
$\implies (x+y)^3+z^3-x^3-y^3=k-(x+y)^3$
$\implies 2(x+y)^3+z^3=k+x^3+y^3$
Combining this with eq(1) will give:
$2(x+y)^3+z^3=k+k-z^3$
$\implies (x+y)^3+z^3=k$
Confusion and my question:-
I just found that $x^3+y^3+z^3=k=(x+y)^3+z^3$
$\implies x^3+y^3+z^3=(x+y)^3+z^3$
What does that supposed to mean? I gave absolutely no restrictions to $x,y,z$ except the assumption in my first step, but that assumption shouldn't really matter.
The equation $x^3+y^3+z^3=(x+y)^3+z^3$
$\implies$ that either $x$ or $y$ is $0$.
Which means that I cannot add three cubes. What does that supposed to mean?!!
Please tell me what I did wrong in the steps.
You made some sign errors.
From the line $$(x+y)^3=k-z^3+3xy(x+y)$$ the next line should be $$ (x+y)^3+z^3=k+3x^2y+3xy^2 $$ Even allowing that error, from the line $$ (x+y)^3+z^3=k-3x^2y-3xy^2 $$ the next line should be $$ (x+y)^3+z^3+x^3+y^3=k-(x+y)^3 $$ An easy way to have found the error on your own is to choose $x,y,z,k$ such that $$ x^3+y^3+z^3=k $$ e.g., $x=1,y=2,z=3,k=36$, and then substitute line-by-line until equality fails on some line.