Zadani su brojevi $A=8x^{3}y$$A=8x^{3}y$ i $B=\frac{1}{2}x^{-3}y^{2}$$B=\frac{1}{2}x^{-3}y^{2}$.

Množimo dva monoma $A = 8x^3y$$A = 8x^3y$ i $B = \frac{1}{2}x^{-3}y^2$$B = \frac{1}{2}x^{-3}y^2$.
Grupiramo koeficijente i potencije s istim bazama:
$A \cdot B = \left(8 \cdot \frac{1}{2}\right) \cdot (x^3 \cdot x^{-3}) \cdot (y \cdot y^2)$$A \cdot B = \left(8 \cdot \frac{1}{2}\right) \cdot (x^3 \cdot x^{-3}) \cdot (y \cdot y^2)$
Zbrajamo eksponente: $x^{3-3} = x^0 = 1$$x^{3-3} = x^0 = 1$ i $y^{1+2} = y^3$$y^{1+2} = y^3$.
Konačni rezultat: $4 \cdot 1 \cdot y^3 = 4y^3$$4 \cdot 1 \cdot y^3 = 4y^3$.
Odgovor: $4y^3$$4y^3$

Potenciramo izraz $B = \frac{1}{2}x^{-3}y^2$$B = \frac{1}{2}x^{-3}y^2$ (odnosno $2^{-1}x^{-3}y^2$$2^{-1}x^{-3}y^2$) na eksponent $-4$$-4$.
Koristimo pravilo $(abc)^n = a^n b^n c^n$$(abc)^n = a^n b^n c^n$ i $(x^m)^n = x^{m\cdot n}$$(x^m)^n = x^{m\cdot n}$:
$B^{-4} = (2^{-1})^{-4} \cdot (x^{-3})^{-4} \cdot (y^2)^{-4}$$B^{-4} = (2^{-1})^{-4} \cdot (x^{-3})^{-4} \cdot (y^2)^{-4}$
$B^{-4} = 2^4 \cdot x^{12} \cdot y^{-8}$$B^{-4} = 2^4 \cdot x^{12} \cdot y^{-8}$
$B^{-4} = 16x^{12}y^{-8}$$B^{-4} = 16x^{12}y^{-8}$
Odgovor: $16x^{12}y^{-8}$$16x^{12}y^{-8}$