Question

Find the monomial if the expression is the cube of the monomial.
-27/64 a to the power of 9 * a to the power of 6

Answer 1

To find the monomial \( x \) such that

\[
x^3 = -\frac{27}{64} a^9 \cdot a^6,
\]

we first simplify the right side of the equation.

Using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \), we combine the exponents on \( a \):

\[
a^9 \cdot a^6 = a^{9+6} = a^{15}.
\]

So, our expression simplifies to:

\[
x^3 = -\frac{27}{64} a^{15}.
\]

Next, we take the cube root of both sides to find \( x \):

\[
x = \sqrt[3]{-\frac{27}{64}} \cdot \sqrt[3]{a^{15}}.
\]

Now, we can simplify each term separately:

1. To simplify \( \sqrt[3]{-\frac{27}{64}} \):
- The cube root of -27 is -3 (as \((-3)^3 = -27\)).
- The cube root of 64 is 4 (as \(4^3 = 64\)).
- Thus, \( \sqrt[3]{-\frac{27}{64}} = -\frac{3}{4} \).

2. To simplify \( \sqrt[3]{a^{15}} \):
- Since the cube root of \( a^{15} \) is \( a^{15/3} = a^5 \).

Putting it all together, we have:

\[
x = -\frac{3}{4} a^5.
\]

Thus, the monomial \( x \) is

\[
\boxed{-\frac{3}{4} a^5}.
\]