If x+ y + z = 11, `x^(2) + y^(2) + z^(2) = 133 and x^(3) + y^(3) + z^(3)` = 881, then the value of `root(8)(xyz)` is :

Options

A. -8
B. -6
C. 6
D. 8

Answer

To solve the problem step by step, we will use the given equations involving \(x\), \(y\), and \(z\):

1. **Given Equations**:
- \(x + y + z = 11\) (1)
- \(x^2 + y^2 + z^2 = 133\) (2)
- \(x^3 + y^3 + z^3 = 881\) (3)

2. **Finding \(xy + yz + zx\)**:
We can use the identity:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \]
Substituting the values from equations (1) and (2):
\[ 11^2 = 133 + 2(xy + yz + zx) \]
\[ 121 = 133 + 2(xy + yz + zx) \]
Rearranging gives:
\[ 2(xy + yz + zx) = 121 - 133 = -12 \]
Thus:
\[ xy + yz + zx = \frac{-12}{2} = -6 \quad (4) \]

3. **Finding \(xyz\)**:
We can use the identity for the sum of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)((x + y + z)^2 - 3(xy + yz + zx)) \]
Substituting the known values:
\[ 881 - 3xyz = 11 \left(11^2 - 3(-6)\right) \]
First, calculate \(11^2 - 3(-6)\):
\[ 11^2 = 121 \]
\[ 3(-6) = -18 \quad \Rightarrow \quad 121 + 18 = 139 \]
Now substituting back:
\[ 881 - 3xyz = 11 \times 139 \]
Calculate \(11 \times 139\):
\[ 11 \times 139 = 1529 \]
Therefore:
\[ 881 - 3xyz = 1529 \]
Rearranging gives:
\[ -3xyz = 1529 - 881 = 648 \]
Thus:
\[ xyz = \frac{-648}{3} = -216 \quad (5) \]

4. **Finding \(\sqrt[8]{xyz}\)**:
We need to find \(\sqrt[8]{xyz}\):
\[ \sqrt[8]{xyz} = \sqrt[8]{-216} \]
Since \(-216\) can be expressed as:
\[ -216 = -1 \times 216 \]
We can separate this:
\[ \sqrt[8]{-216} = \sqrt[8]{-1} \times \sqrt[8]{216} \]
The eighth root of \(-1\) is not a real number, but we can focus on the magnitude:
\[ \sqrt[8]{216} = \sqrt[8]{6^3} = 6^{3/8} \]
Since we need the real part, we can express it as:
\[ \sqrt[8]{216} = 6^{3/8} \]

5. **Final Answer**:
The value of \(\sqrt[8]{xyz}\) is:
\[ \sqrt[8]{-216} = -6^{3/8} \]