1. If $x-y=12$, find the value of $x^{3}-y^{3}-36 x y$.
(1 mark)
If $x-y=12$, find the value of $x^{3}-y^{3}-36 x y$.
(1 mark)
1. 1728
1. We have $x^{3}-y^{3}-36 x y=(x-y)\left(x^{2}+x y+y^{2}\right)-36 x y$
$$
\begin{array}{l}
=12\left(x^{2}+x y+y^{2}\right)-36 x y \\
=12\left(x^{2}+x y+y^{2}-3 x y\right) \\
=12(x-y)^{2} \\
=12^{3} \\
=1728
\end{array}
$$

Remark. One may guess the answer by substituting suitable values of $x$ and $y$.