For those who don't know Part 7 (Determinants)

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

By taking common factors $(x-y)$ from ${ C }_{ 1 }$ , y from ${ C }_{ 2 }$ and x from ${ C }_{ 3 }$ .

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

One of the proprieties of determinant is when two rows or two columns are equal ,the determinant equals to $0$

$\because { C }_{ 1 }={ C }_{ 3 }\\ \therefore xy(x-y)[0]=\boxed { 0 }$

• If we look closely enough, we can figure it out that the first column and the third column are equal.
• The third column is the result of the first column times $( \frac{x}{x - y} )$ .
• As result, when two or more columns are equal, the determinant is zero.