Suppose a,b,c are distinct real numbers. Let
`P(x)=|(0,x^(3)-a,x^(4)-b),(x^(3)+a,0,x^(5)+c),(x^(4)+b,x^(5)-c,0)|`
A value of x satisfying `P(x)=0` is

Options

A. `-(a+b+c)`
B. `a+b+c`
C. `a+b-c`
D. 0

Answer

To solve the problem, we need to evaluate the determinant \( P(x) = \begin{vmatrix} 0 & x^3 - a & x^4 - b \\ x^3 + a & 0 & x^5 + c \\ x^4 + b & x^5 - c & 0 \end{vmatrix} \) and find the values of \( x \) for which \( P(x) = 0 \).

### Step 1: Write down the determinant
We have:
\[ P(x) = \begin{vmatrix} 0 & x^3 - a & x^4 - b \\ x^3 + a & 0 & x^5 + c \\ x^4 + b & x^5 - c & 0 \end{vmatrix} \]

### Step 2: Expand the determinant using the first row
Using the first row to expand the determinant:
\[ P(x) = 0 \cdot \begin{vmatrix} 0 & x^5 + c \\ x^5 - c & 0 \end{vmatrix} - (x^3 - a) \cdot \begin{vmatrix} x^3 + a & x^5 + c \\ x^4 + b & 0 \end{vmatrix} + (x^4 - b) \cdot \begin{vmatrix} x^3 + a & 0 \\ x^4 + b & x^5 - c \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
1. For the first 2x2 determinant:
\[ \begin{vmatrix} x^3 + a & x^5 + c \\ x^4 + b & 0 \end{vmatrix} = 0 - (x^3 + a)(x^4 + b) = -(x^3 + a)(x^4 + b) \]

2. For the second 2x2 determinant:
\[ \begin{vmatrix} x^3 + a & 0 \\ x^4 + b & x^5 - c \end{vmatrix} = (x^3 + a)(x^5 - c) - 0 = (x^3 + a)(x^5 - c) \]

### Step 4: Substitute back into the determinant expression
Substituting these results back into the expression for \( P(x) \):
\[ P(x) = -(x^3 - a)(-(x^3 + a)(x^4 + b)) + (x^4 - b)((x^3 + a)(x^5 - c)) \]
\[ = (x^3 - a)(x^3 + a)(x^4 + b) + (x^4 - b)(x^3 + a)(x^5 - c) \]

### Step 5: Set \( P(x) = 0 \)
We need to find \( x \) such that:
\[ (x^3 - a)(x^3 + a)(x^4 + b) + (x^4 - b)(x^3 + a)(x^5 - c) = 0 \]

### Step 6: Analyze the factors
Notice that if \( x^3 = a \) or \( x^3 = -a \), then \( P(x) \) will be zero. Thus, we can find values of \( x \):
1. If \( x^3 = a \), then \( x = a^{1/3} \).
2. If \( x^3 = -a \), then \( x = -a^{1/3} \).

### Conclusion
Therefore, a value of \( x \) satisfying \( P(x) = 0 \) is:
\[ x = a^{1/3} \quad \text{or} \quad x = -a^{1/3} \]