If we take x+y = 2 and Cube it,
We have x3+y3+6xy = 8
So any I think any two integers which satisfy x + y = 2, satisfy the equation. Please Correct me!
\hspace{-16}$Find all Integer solution $\mathbf{(x,y)}$ in $\mathbf{x^3+y^3+6xy=8}$
If we take x+y = 2 and Cube it,
We have x3+y3+6xy = 8
So any I think any two integers which satisfy x + y = 2, satisfy the equation. Please Correct me!
hmm you are right.
we can also have:
x3+y3+(-2)3=3*x*y*(-2)
i.e. a3+b3+c3=3abc
then we have a+b+c=0