Consider a 3 ×3matrix A whose (i,j)^-th element, a_i,j = (i-j)³ . Then the matrix A will be

The correct option is skew-symmetric.

To determine the nature of the matrix $A$, let us analyze its elements. The matrix $A$ is a $3\times 3$ matrix where the element in the $i$-th row and $j$-th column is given by the formula:
$a_{i,j}={(i-j)}^3$

Recall the definition of a skew-symmetric matrix. A matrix $A$ is skew-symmetric if its transpose is equal to its negative, which means:
$A^T=-A$
In terms of individual elements, this condition is written as:
$a_{j,i}=-a_{i,j}$
for all indices $i$ and $j$.

Let us check if this property holds for the elements of matrix $A$:
$a_{j,i}={(j-i)}^3$

We can factor out $-1$ from the term inside the parenthesis:
$j-i=-(i-j)$

Now, substituting this back into the expression for $a_{j,i}$ gives:
$a_{j,i}={[-(i-j\left)\right]}^3$
Since the cube of $-1$ is $-1$ (i.e., ${(-1)}^3=-1$), we get:
$a_{j,i}=-{(i-j)}^3$

Using the definition of $a_{i,j}$, we have:
$a_{j,i}=-a_{i,j}$

Additionally, let us evaluate the diagonal elements where $i=j$:
$a_{i,i}={(i-i)}^3=0^3=0$
All diagonal elements are zero, which is a key property of a skew-symmetric matrix.

Since the relation $a_{j,i}=-a_{i,j}$ holds true for all $i$ and $j$, the matrix $A$ is skew-symmetric.