If (x + 6y) = 8, and xy = 2, where x > 0, what is the value of (x³ + 216y³)?

To find the value of the expression $x^3+216y^3$, we can use algebraic identities.

First, let's identify the relationship between the terms in the expression we want to find and the given equations:
$x+6y=8$
$xy=2$

Notice that the term $216y^3$ can be rewritten as a perfect cube:
$216y^3={\left(6y\right)}^3$

Thus, the expression we want to evaluate is:
$x^3+{\left(6y\right)}^3$

We can use the standard algebraic identity for the sum of two cubes:
$a^3+b^3={(a+b)}^3-3ab(a+b)$

By letting $a=x$ and $b=6y$, we can write:
$x^3+{\left(6y\right)}^3={(x+6y)}^3-3\left(x\right)\left(6y\right)(x+6y)$

Simplifying the product term on the right side gives:
$x^3+216y^3={(x+6y)}^3-18xy(x+6y)$

Now, substitute the known values $x+6y=8$ and $xy=2$ into the equation:
$x^3+216y^3=8^3-18\cdot \left(2\right)\cdot \left(8\right)$

Calculate the terms individually:
$8^3=512$
$18\cdot 2\cdot 8=36\cdot 8=288$

Subtract the two values to get the final answer:
$x^3+216y^3=512-288=224$

Therefore, the correct option is 224.