If `Delta=|(a,p,x),(b,q,y),(c,r,z)|=16` then `Delta_(1)=|(p+x,a+x,a+p),(q+y,b+y,b+q),(r+z,c+z,c+r)|=32`

Explanation

To solve the problem, we need to evaluate the determinant \( \Delta_1 \) given that \( \Delta = |(a,p,x),(b,q,y),(c,r,z)| = 16 \).

### Step-by-Step Solution:

1. **Understanding the Determinant**:
We start with the determinant \( \Delta = |(a,p,x),(b,q,y),(c,r,z)| \). We know that \( \Delta = 16 \).

2. **Transforming the Columns**:
We need to evaluate \( \Delta_1 = |(p+x,a+x,a+p),(q+y,b+y,b+q),(r+z,c+z,c+r)| \).
We can apply the transformation to the first column:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ \Delta_1 = |(2p + x + a, a + x, a + p), (2q + y + b, b + y, b + q), (2r + z + c, c + z, c + r)| \]

3. **Factoring Out Common Terms**:
We can factor out a 2 from the first column:
\[ \Delta_1 = 2 |(p + x + a, a + x, a + p), (q + y + b, b + y, b + q), (r + z + c, c + z, c + r)| \]

4. **Further Transformations**:
Now, we apply the transformation:
\[ C_1 \rightarrow C_1 - C_2 \quad \text{and} \quad C_2 \rightarrow C_2 - C_3 \]
This results in:
\[ \Delta_1 = 2 |(p, a + x - (a + p), a + p), (q, b + y - (b + q), b + q), (r, c + z - (c + r), c + r)| \]
Simplifying gives:
\[ = 2 |(p, x - p, a + p), (q, y - q, b + q), (r, z - r, c + r)| \]

5. **Identifying Identical Rows**:
We can see that if we split this determinant into two parts, we will have:
\[ 2 \times pqr \times |(1, 1, 1), (1, 1, 1), (1, 1, 1)| \]
Since the last two rows are identical, the determinant evaluates to zero.

6. **Final Evaluation**:
Thus, we are left with:
\[ \Delta_1 = 2 \times 16 = 32 \]

### Conclusion:
Therefore, the statement that \( \Delta_1 = 32 \) is true.