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

The correct answer is 64.

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

We want to find the value of the expression:
$x^3-27y^3-36xy$

Step 1: Rearrange the given equation
Subtract $3y$ from both sides of the equation:
$x-3y=4$

Step 2: Cube both sides of the rearranged equation
Using the algebraic identity for the cube of a binomial, ${(a-b)}^3=a^3-b^3-3ab(a-b)$, let $a=x$ and $b=3y$:

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

Expanding the left side:
$x^3-{\left(3y\right)}^3-3\left(x\right)\left(3y\right)(x-3y)=64$

Simplify the terms:
$x^3-27y^3-9xy(x-3y)=64$

Step 3: Substitute the value of $(x-3y)$
From Step 1, we know that $x-3y=4$. Substitute $4$ into the expression:

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

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

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