To solve the problem, we start with the given information and the relationship between [tex]\( y \)[/tex] and [tex]\( x \)[/tex].
1. Understanding Direct Proportion:
- We know that [tex]\( y \)[/tex] is directly proportional to the cube root of [tex]\( (x + 3) \)[/tex]. This can be mathematically represented as:
[tex]\[
y = k \cdot \sqrt[3]{x + 3}
\][/tex]
where [tex]\( k \)[/tex] is the constant of proportionality.
2. Given Values:
- When [tex]\( x = 5 \)[/tex], [tex]\( y = \frac{2}{3} \)[/tex].
3. Finding the Constant of Proportionality [tex]\( k \)[/tex]:
- Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = \frac{2}{3} \)[/tex] into the equation [tex]\( y = k \cdot \sqrt[3]{x + 3} \)[/tex]:
[tex]\[
\frac{2}{3} = k \cdot \sqrt[3]{5 + 3}
\][/tex]
- Simplify inside the cube root:
[tex]\[
\frac{2}{3} = k \cdot \sqrt[3]{8}
\][/tex]
- Since the cube root of 8 is 2, we have:
[tex]\[
\frac{2}{3} = k \cdot 2
\][/tex]
- Solve for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{\frac{2}{3}}{2} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}
\][/tex]
4. Using [tex]\( k \)[/tex] to Find [tex]\( y \)[/tex] When [tex]\( x = 24 \)[/tex]:
- Substitute [tex]\( x = 24 \)[/tex] and [tex]\( k = \frac{1}{3} \)[/tex] back into the equation:
[tex]\[
y = \frac{1}{3} \cdot \sqrt[3]{24 + 3}
\][/tex]
- Simplify inside the cube root:
[tex]\[
y = \frac{1}{3} \cdot \sqrt[3]{27}
\][/tex]
- Since the cube root of 27 is 3, we have:
[tex]\[
y = \frac{1}{3} \cdot 3 = 1
\][/tex]
Therefore, when [tex]\( x = 24 \)[/tex], the value of [tex]\( y \)[/tex] is [tex]\( \boxed{1} \)[/tex].
