Question

If $x^3+y^3+1=3xy$, where $x\ne y$ determine the value of $x+y+1$.

Answer 1

Let $x+y=t$

Cubing on both sides, we get

$\Rightarrow (x+y{)}^3=t^3$

$\Rightarrow x^3+y^3+3xy(x+y)=t^3$

$\Rightarrow x^3+y^3+3xyt=t^3$

$\Rightarrow x^3+y^3-t^3=-3xyt$

Substitute $t=-1$, we get

$\Rightarrow x^3+y^3-(-1{)}^3=-3xy(-1)$

$\Rightarrow x^3+y^3+1=3xy$ which is the given equation

Therefore, $x+y=t$

$\Rightarrow x+y=-1$

$\Rightarrow x+y+1=0$