If $x=3y+4$, then what is the value of $(x^3-27y^3-36xy)$?

Given equation:
$$x=3y+4$$

Rearranging the terms, we get:

$$x-3y=4$$

Now, cube both sides of the equation:

$$(x-3y{)}^3=4^3$$

Using the algebraic identity $(a-b{)}^3=a^3-b^3-3ab(a-b)$:

$$x^3-(3y{)}^3-3(x)(3y)(x-3y)=64$$

$$x^3-27y^3-9xy(x-3y)=64$$

Substitute the value $x-3y=4$ into the equation:

$$x^3-27y^3-9xy(4)=64$$

$$x^3-27y^3-36xy=64$$

Therefore, the value of $(x^3-27y^3-36xy)$ is $64$.