For those who don't know Part 6 (Determinants)

$X=\left| \begin{matrix} x & y & 1 \\ { x }^{ 2 } & { y }^{ 2 } & x+y \\ { x }^{ 3 } & { y }^{ 3 } & { x }^{ 2 }+xy+{ y }^{ 2 } \end{matrix} \right|$

$C_{1} \to C_{1} - C_{2}$

$X = \left| \begin{matrix} (x - y) & y & 1 \\ (x - y)(x+y) & { y }^{ 2 } & x+y \\ ( x - y )({ x }^{ 2 }+xy+{ y }^{ 2 })& { y }^{ 3 } & { x }^{ 2 }+xy+{ y }^{ 2 } \end{matrix} \right|$

$\boxed{Using ~ x^2 - y^2 = (x - y)(x+y) , x^3 - y^3 = (x - y)(x^2 + xy + y^2)}$

$X = (x - y) \left| \begin{matrix} 1 & y & 1 \\ x+y & { y }^{ 2 } & x+y \\ ({ x }^{ 2 }+xy+{ y }^{ 2 })& { y }^{ 3 } & { x }^{ 2 }+xy+{ y }^{ 2 } \end{matrix} \right|$

$C_{1} \to C_{1} - C_{3}$

$X = (x - y) \left| \begin{matrix} 0 & y & 1 \\ 0 & { y }^{ 2 } & x+y \\ 0 & { y }^{ 3 } & { x }^{ 2 }+xy+{ y }^{ 2 } \end{matrix} \right|$

Now expand, along 1st column

Thus $X =0$

I'm not sure that any special tricks are needed. This one can be done with a formal algebra system: