Question

$$-108 { a }^{ 3 } { b }^{ 3 } cBY18abc$$

Answer

$$-1944*cBY*a^4*b^4*c$$

Solution


Take out the constants.
\[-(108\times 18){a}^{3}a{b}^{3}bccBY\]
Simplify  \(108\times 18\)  to  \(1944\).
\[-1944{a}^{3}a{b}^{3}bccBY\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[-1944{a}^{3+1}{b}^{3+1}ccBY\]
Simplify  \(3+1\)  to  \(4\).
\[-1944{a}^{4}{b}^{4}ccBY\]
Regroup terms.
\[-1944cBY{a}^{4}{b}^{4}c\]