Question

1. simplify: $(3a^{3}b^{5}c)^{2}(4ab^{2}c^{3})(2ab^{-5})^{0}$

Explanation:

Step1: Simplify the power of a product

First, we simplify \((3a^{3}b^{5}c)^{2}\) using the power of a product rule \((xy)^n = x^n y^n\).
\[

$$\begin{align*} (3a^{3}b^{5}c)^{2}&=3^{2}(a^{3})^{2}(b^{5})^{2}c^{2}\\ &=9a^{6}b^{10}c^{2} \end{align*}$$

\]

Step2: Simplify the zero exponent

Next, we simplify \((2ab^{-5})^{0}\). Any non - zero number to the power of 0 is 1, so \((2ab^{-5})^{0}=1\) (assuming \(2ab^{-5}
eq0\), which is true for non - zero \(a\) and \(b\)).

Step3: Multiply the remaining terms

Now we multiply \(9a^{6}b^{10}c^{2}\), \(4ab^{2}c^{3}\) and 1. Using the product rule of exponents \(x^m\times x^n=x^{m + n}\) for the same base:
\[

$$\begin{align*} &9a^{6}b^{10}c^{2}\times4ab^{2}c^{3}\times1\\ =&(9\times4)\times(a^{6}\times a)\times(b^{10}\times b^{2})\times(c^{2}\times c^{3})\\ =&36\times a^{6 + 1}\times b^{10+2}\times c^{2 + 3}\\ =&36a^{7}b^{12}c^{5} \end{align*}$$

\]

Answer:

\(36a^{7}b^{12}c^{5}\)