Diophantine Equataion of Degree 6
=================================
Another from YouTube, solved by Michael Penn:
Find 2 integers x,y such that

  x⁶+x³y = y³+2y²

Let us move x³y to the right side:

  x⁶ = y³+2y²-x³y
     = y(y²+2y-x³)

If y=0, then:

  x⁶ = 0(0²+2·0²-0·x³) = 0 ==> x=0

So, our first solution is:

  (x,y)=(0,0)

Otherwise y|x⁶

Does it mean that y|x³ ?

Not sure, but there is a rational number r, such that x³=ry since both y
and x³ are integers.

Let us plug it into our original equation:

       (ry)²+(ry)y = y³+2y² ==>

  ==>  r²y²+ry²=y³+2y²

Since y≠0, we can divide both sides by y²:

      r²+r=y+2 ==>
  ==> y=r²+r-2 ==>
  ==> x³=ry=r³+r²-2r

Now, according to the rational root theorem, if for an integer value of y, 
the equation:

   r²+r-(y+2)=0

has a rational solution, then r is an integer.

Now, if r=0, then

  x=0

and

  y =r²+r-2=-2

So, our second solution is 

  (x,y)=(0,-2)
-------------------------
If r≠0, recall that

  x³=r³+r²-2r

Now, let us see when 

  r³<r³+r²-2r<(r+1)³

which means that r³+r²-2r is not a perfect cube

It's clear that for r>0

  r³+r²-2r<r³+3r²+3r+1=(r+1)³

and

      r³<r³+r²-2r ==>
  ==> r²-2r>0 ==>
  ==> r(r-2)>0

which means that:

  r>2 ==> r³<r³+r²-2r<(r+1)³

Thusm there are no solutions for r>2

Let us check r=1

  r=1 ==> x³=1³+1²-2=0 and y=1²+1-2=0

That's a contradication to our assumption that y≠0

Let us check r=2

  r=2 ==> x³=2³+2²-4=8=2³ and y=2²+2-2=4

So, our third solution is

  (x,y)=(2,4)

-------------------
If r<0, we have already shown that r³+r²-2r>r³

Let us check when r³+r²-2r<(r+1)³

      r³+r²-2r<(r+1)³ ==>
  ==> r³+r²-2r<r³+3r²+3r+1 ==>
  ==> -2r²-5r-1<0 ==>
  ==> 2r²+5r+1>0 ==>
  ==> 16r²+40r+8>0 ==>
  ==> (4r)²+2·4r·5+25-17>0 ==>
  ==> (4r+5)²>17

For sure,

  4r+5>5 satisfies that, and that means that
  4r>0 ==> r>0

  which contradicts our assumption that r<0

And

  4r+5<-5 satisfies our inequality as well, and that means that:
  4r<-10 ==> r<-2.5

  which means that if r<-2:

  r³<r³+r²-2r<(r+1)³

  thus, r³+r²-2r is not a perfect cube if r<-2

Let us check values of x³ if r is an integer between -2 and -1

  r=-2==>x³=(-2)³+(-2)²-2(-2)=-8+4+4=0 ==> y=0 - again?

  r=-1==>x³=(-1)³+(-1)²-2(-1)=-1+1+2=2 - not a perfect cube.

Thus, our solutions are:
  (x,y)=(0,-2)
  (x,y)=(0,0)
and
  (x,y)=(2,4)