x^3+y^3+3x^2*y^2=x^3*y^3
=>x^3+y^3=x^3*y^3-3*x^2*y^2
=> x^3*y^3-(x+y)^3=3*x^2*y^2
=> (xy-x-y)(x^2+y^2+(x+y)^2+xy(x+y)=3xy(xy-x-y)
=> (xy-x-y)(x^2*y^2+(x+y)^2+xy(x+y)-3xy)=0
So, either xy-x-y=0
Or, x^2*y^2+(x+y)^2+xy(x+y)-3xy=0
CASE 1
xy-x-y=0, gives:
xy=x+y, so that:
1/x+1/y=1 .........(1)
CASE 2
x^2*y^2+(x+y)^2+xy(x+y)-3xy=0
We first simplify the lhs
x^2*y^2+(x+y)^2+xy(x+y)-3xy
=x^2*y^2+xy(x+y)+(x+y)^2-4xy+xy
= x^2*y^2+xy(x+y+1)+(x-y)^2
=x^2*y^2+xy(x+y)+xy+(x-y)^2
=(xy+(x+y)/2)^2- ((x+y)/2)^2+xy+(x-y)^2
= (xy+(x+y)/2)^2 -((x-y)/2)^2+(x-y)^2
=(xy+(x+y)/2)^2 +(3 /4)*(x-y)^2........(#)
Now, the above (#) is a sum of a square and positive rational multiple of a square.
Thus, (#) is zero only when the individual expressions themselves equal zero.
Therefore, we must have:
xy+(x+y)/2=0 and, x-y=0
So, x=y(=n, say). This gives:
n^2+n=0, so that: n=0,-1
Consequently, (x,y)=(0,0), (-1,-1)
(x,y)=(-1,-1), gives, 1/x+1/y=-2
(x y)=(0,0), gives 1/x+1/y as undefined.
Hence, the FORMAL ANSWER is:
1/x+1/y = 1, -2
Or, 1/x+1/y is undefined.
It may be NOTED that this is a fully algebraic solution to the given problem, WITH NO GRAPHING.
Edited on December 17, 2021, 10:24 pm
Edited on December 18, 2021, 2:14 am
Solution
