x^2+xy+y^2=(x+y+3)^3/27,find all(x,y) It is obvious that $x+y$ is divisible by $3$ . Put $x+y$ = $3k$ . So, the equation becomes, $$(3k)^2 - x(3k-x) = (k+1)^3 $$ => $$ x^2 - 3kx + 9k^2 = k^3 + 3k^2 + 3k +1 $$ => $$ x^2 - 3kx - k^3 + 6k^2 - 3k -1 = 0 $$ As we have to find the integer roots of this e...