Beginning Algebra With Applications, Chapter 7, 7.2, Section 7.2, Problem 88

Simplify $\displaystyle 9x^3 (3x^2 y)^2 - x(x^3 y)^2$


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\begin{equation}
\begin{aligned}

9x^3 (3x^2 y)^2 - x(x^3 y)^2 =& 9x^3 (3)^2 x^4 y^2 - x \left( x^6 y^2 \right)
&& \text{Multiply each exponent in $3x^2 y$ and in $x^3 y$ by the exponent outside the parentheses}
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=& (9x^3) \left( 9x^4 y^2 \right) - x \left( x^6 y^2 \right)
&& \text{Simplify } (3)^2
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=& (9 \cdot 9) \left( x^3 \cdot x^4 \right) y^2 - (x \cdot x^6) y^2
&& \text{Use Properties of Multiplication to rearrange and group factors}
\\
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=& 81x^7 y^2 - x^7 y^2
&& \text{Multiply variables with the same base by adding the exponents}
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=& 80 x^7 y^2
&&

\end{aligned}
\end{equation}
$