Question

Simplify evaluate the numerical bases (4^2b^3c-5)(4^-1b^-4c^8)

Answer 1

Given expression 
$(4^2b^3c^{-5})(4^{-1}b^{-4}c^8)$

Rewriting each term
$4^2 = 16$
$b^3 = b^3$
$c^{-5}= \frac{1}{c^5}$
$4^{-1} = \frac{1}{4}$
$b^{-4} = \frac{1}{b^4}$
$c^8 = c^8$

Put them back together
$(16)(b^3)( \frac{1}{c^5} )( \frac{1}{4})( \frac{1}{b^4})(c^8)$
$( \frac{16}{4})( \frac{b^3}{b^4})( \frac{c^8}{c^5} )$
$4(b^{3-4})(c^{8-5})$
$4b^{-1}c^3$
$\frac{4c^3}{b}$