Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(-\frac{4}{7} a^{2} b\right) \times\left(-\frac{2}{3} b^{2} c\right) \times\left(-\frac{7}{6} c^{2} a\right)$
To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$.
We have:
$\left(-\frac{4}{7} a^{2} b\right) \times\left(-\frac{2}{3} b^{2} c\right) \times\left(-\frac{7}{6} c^{2} a\right)$
$=\left\{\left(-\frac{4}{7}\right) \times\left(-\frac{2}{3}\right) \times\left(-\frac{7}{6}\right)\right\} \times\left(a^{2} \times a\right) \times\left(b \times b^{2}\right) \times\left(c \times c^{2}\right)$
$=\left\{\left(-\frac{4}{7}\right) \times\left(-\frac{2}{3}\right) \times\left(-\frac{7}{6}\right)\right\} \times\left(a^{2+1}\right) \times\left(b^{1+2}\right) \times\left(c^{1+2}\right)$
$=-\frac{4}{9} a^{3} b^{3} c^{3}$
$\because$ The expression doesn't consist of the variables $x$ and $y$.
$\therefore$ The result cannot be verified for $x=1$ and $y=2$.
Thus, the answer is $-\frac{4}{9} a^{3} b^{3} c^{3}$.