Let `f(x)=|(1,x,x^(2)),(x,x^(2),1),(x^(2),1,x)|` then `|f(2.9)^(1//3)|=`___________

Answer

To solve the problem, we need to compute the determinant \( f(x) = |(1, x, x^2), (x, x^2, 1), (x^2, 1, x)| \) and then evaluate \( |f(2.9)|^{1/3} \).

### Step 1: Calculate the determinant \( f(x) \)

The determinant can be calculated using the formula for a 3x3 matrix:

\[ f(x) = |(1, x, x^2), (x, x^2, 1), (x^2, 1, x)| \]

Using the determinant formula for a 3x3 matrix:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

For our matrix:

\[ A = \begin{pmatrix} 1 & x & x^2 \\ x & x^2 & 1 \\ x^2 & 1 & x \end{pmatrix} \]

We can assign:

- \( a = 1, b = x, c = x^2 \)
- \( d = x, e = x^2, f = 1 \)
- \( g = x^2, h = 1, i = x \)

Now, we compute the determinant:

\[ f(x) = 1 \cdot (x^2 \cdot x - 1 \cdot 1) - x \cdot (x \cdot x - 1 \cdot x^2) + x^2 \cdot (x \cdot 1 - x^2 \cdot 1) \]

Calculating each term:

1. \( 1 \cdot (x^3 - 1) = x^3 - 1 \)
2. \( -x \cdot (x^2 - x^2) = 0 \)
3. \( x^2 \cdot (x - x^2) = x^2(x - x^2) = x^3 - x^4 \)

Putting it all together:

\[ f(x) = (x^3 - 1) + 0 + (x^3 - x^4) = 2x^3 - x^4 - 1 \]

### Step 2: Evaluate \( f(2.9) \)

Now we substitute \( x = 2.9 \):

\[ f(2.9) = 2(2.9)^3 - (2.9)^4 - 1 \]

Calculating \( (2.9)^3 \) and \( (2.9)^4 \):

\[ (2.9)^3 = 24.389 \quad \text{and} \quad (2.9)^4 = 70.561 \]

Now substituting back:

\[ f(2.9) = 2 \cdot 24.389 - 70.561 - 1 \]

Calculating:

\[ f(2.9) = 48.778 - 70.561 - 1 = 48.778 - 71.561 = -22.783 \]

### Step 3: Calculate \( |f(2.9)|^{1/3} \)

Now we need to find the absolute value and then take the cube root:

\[ |f(2.9)| = |-22.783| = 22.783 \]

Finally, we compute:

\[ |f(2.9)|^{1/3} = (22.783)^{1/3} \]

Calculating the cube root:

\[ |f(2.9)|^{1/3} \approx 2.843 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{2.843} \]