**Hint:** As per the given polynomial, and the condition given, we just need to replace the given condition into \[{{\text{x}}^{\text{3}}}{\text{-8}}{{\text{y}}^{\text{3}}}{\text{ - 36xy - 216,}}\] this meansl means directly replace the value of x in the given polynomial and solve to get the required answer.

**Complete step by step answer:**

The given polynomial is \[{{\text{x}}^{\text{3}}}{\text{ - 8}}{{\text{y}}^{\text{3}}}{\text{ - 36xy - 216,}}\]and the condition is \[{\text{x = 2y + 6}}\]

So, replacing the value of x in to the polynomial we get,

\[

\Rightarrow {{\text{x}}^{\text{3}}}{\text{ - 8}}{{\text{y}}^{\text{3}}}{\text{ - 36xy - 216}} \\

{\text{as,x = 2y + 6}} \\

\Rightarrow {\text{(2y + 6}}{{\text{)}}^{\text{3}}}{\text{ - 8}}{{\text{y}}^{\text{3}}}{\text{ - 36(2y + 6)y - 216}} \\

\]

Now, using the formula of \[{{\text{(a + b)}}^{\text{3}}}{\text{ = }}{{\text{a}}^{\text{3}}}{\text{ + }}{{\text{b}}^{\text{3}}}{\text{ + 3ab(a + b)}}\]and simplifying the equation further,

\[

{\text{ = (2y}}{{\text{)}}^{\text{3}}}{\text{ + (6}}{{\text{)}}^{\text{3}}}{\text{ + 3(2y)(6)(2y + 6) - 8}}{{\text{y}}^{\text{3}}}{\text{ - 36(2y + 6)y - 216}} \\

{\text{ = 8}}{{\text{y}}^{\text{3}}}{\text{ + 216 + 36(2y + 6)y - 8}}{{\text{y}}^{\text{3}}}{\text{ - 36(2y + 6)y - 216}} \\

{\text{ = 0}} \\

\]

Hence, **zero is our correct answer.**

**Note:** In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. Polynomials are of different types. Namely, Monomial, Binomial, and Trinomial. A monomial is a polynomial with one term. A binomial is a polynomial with two, unlike terms.

**Additional information:** A polynomial function is a function that involves only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc. A polynomial that, when evaluated over each in the domain of definition, results in the same value. The simplest example is for and. a constant