If `x. y, z` are positive real numbers such that `x^2+y^2+z^2=27,` then `x^3+y^3+z^3` has

Options

A. minimum value 81
B. maximum value 81
C. minimum value 27
D. maximum value 27

Answer

To find the minimum value of \( x^3 + y^3 + z^3 \) given that \( x^2 + y^2 + z^2 = 27 \) and \( x, y, z \) are positive real numbers, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply the AM-GM Inequality on \( x^2, y^2, z^2 \)**:
\[ \frac{x^2 + y^2 + z^2}{3} \geq \sqrt[3]{x^2 y^2 z^2} \]
Given that \( x^2 + y^2 + z^2 = 27 \), we can substitute this into the inequality:
\[ \frac{27}{3} \geq \sqrt[3]{x^2 y^2 z^2} \]
Simplifying this gives:
\[ 9 \geq \sqrt[3]{x^2 y^2 z^2} \]

2. **Cubing Both Sides**:
\[ 9^3 \geq x^2 y^2 z^2 \]
This simplifies to:
\[ 729 \geq x^2 y^2 z^2 \]

3. **Taking the Square Root**:
\[ 27 \geq xyz \]
Thus, we have established that \( xyz \leq 27 \).

4. **Apply the AM-GM Inequality on \( x^3, y^3, z^3 \)**:
\[ \frac{x^3 + y^3 + z^3}{3} \geq \sqrt[3]{x^3 y^3 z^3} \]
This can be rewritten as:
\[ x^3 + y^3 + z^3 \geq 3 \sqrt[3]{x^3 y^3 z^3} \]

5. **Substituting \( xyz \)**:
Since \( xyz \leq 27 \), we can express \( x^3 y^3 z^3 \) as:
\[ x^3 y^3 z^3 = (xyz)^3 \leq 27^3 = 19683 \]

6. **Finding the Minimum Value**:
Therefore, we can substitute back into our inequality:
\[ x^3 + y^3 + z^3 \geq 3 \sqrt[3]{(xyz)^3} \geq 3 \cdot 27 = 81 \]

7. **Conclusion**:
The minimum value of \( x^3 + y^3 + z^3 \) is \( 81 \).

### Final Answer:
The minimum value of \( x^3 + y^3 + z^3 \) is \( 81 \).