x varies directly as the cube of y, x is 32 when y is 4, then what is the value of y when x is 108 ?

Answer

To solve the problem where \( x \) varies directly as the cube of \( y \), we can follow these steps:

### Step 1: Establish the relationship
Since \( x \) varies directly as the cube of \( y \), we can express this relationship as:
\[ x = k \cdot y^3 \]
where \( k \) is the constant of proportionality.

### Step 2: Find the constant \( k \)
We know that \( x = 32 \) when \( y = 4 \). We can substitute these values into the equation to find \( k \):
\[ 32 = k \cdot (4)^3 \]
Calculating \( (4)^3 \):
\[ (4)^3 = 64 \]
Now substituting back:
\[ 32 = k \cdot 64 \]
To find \( k \), we divide both sides by 64:
\[ k = \frac{32}{64} = \frac{1}{2} \]

### Step 3: Use the value of \( k \) to find \( y \) when \( x = 108 \)
Now we need to find the value of \( y \) when \( x = 108 \). We substitute \( k \) into the original equation:
\[ 108 = \frac{1}{2} \cdot y^3 \]
To eliminate the fraction, multiply both sides by 2:
\[ 2 \cdot 108 = y^3 \]
Calculating the left side:
\[ 216 = y^3 \]

### Step 4: Solve for \( y \)
To find \( y \), we take the cube root of both sides:
\[ y = \sqrt[3]{216} \]
Since \( 216 = 6^3 \):
\[ y = 6 \]

### Final Answer
The value of \( y \) when \( x = 108 \) is:
\[ \boxed{6} \]