Solving a Cubic Equation with Two Variables?

This math Olympiad tutorial video shows you how to solve a special cubic equation x^3+y^3+3xy=1 with two variables x and y. The main techniques (tricks) include polynomial factoring, sum of cubes formula, difference of cubes formula, perfect square formula, and the properties of real numbers.

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[prabhudasmandal]

The equn is (x+y)^3--3xy(x+y)+3xy--1=0.if we take x+y=a, xy=b then it is a^3--3ab+3b--1=0 or regrouping it is (a^3--1^3)--3b(a--1)=0 or (a--1)(a^2+a+1)--3b(a--1)=0 or (a--1)(a^2+a+1--3b)=0.so, a=1.now, consider 2nd part as a quadratic one where a=1, b=1, c={1--3b) and we get a=--1+--√{12b--3)/2.here b=1 gives a=, _1+--√9/2 or( --1+--3)/2 i.e(3--1)/2 or(--3--1)/2 =1, --2 i.e x+y=1, --2.ans.

[nemanjalazarevic]

Couldnt we have just assumed:
(x+y)³=x³+y³+3xy
Expand
x³+y³+3xy(x+y)=x³+y³+3xy
Remove common variables
3xy(x+y)=3xy
x+y=1
Any reason why that is invalid?