If p = 6, q = −3, then the value of p³ – 3p² + 3p + 3q + 3q² + q³ is:

The correct option is 117.

We are given the following values for the variables:
$p=6$
$q=-3$

We need to find the value of the algebraic expression:
$E=p^3-3p^2+3p+3q+3q^2+q^3$

Method 1: Direct Substitution

First, calculate each term involving $p$ using $p=6$:
$p^3=6^3=216$
$-3p^2=-3\left(6^2\right)=-3\left(36\right)=-108$
$3p=3\left(6\right)=18$

Next, calculate each term involving $q$ using $q=-3$:
$3q=3(-3)=-9$
$3q^2=3\left(\right(-3{)}^2)=3(9)=27$
$q^3=(-3{)}^3=-27$

Now, add all the calculated values together:
$E=216-108+18-9+27-27$

Simplifying step-by-step:
$E=(216-108)+18-9+(27-27)$
$E=108+18-9+0$
$E=126-9$
$E=117$

Method 2: Using Algebraic Identities

We can rewrite the expression using algebraic identities for cubic terms:
Recall that $(p-1{)}^3=p^3-3p^2+3p-1$ and $(q+1{)}^3=q^3+3q^2+3q+1$.

Adding these two expansions:
$(p-1{)}^3+(q+1{)}^3=(p^3-3p^2+3p-1)+(q^3+3q^2+3q+1)$
$(p-1{)}^3+(q+1{)}^3=p^3-3p^2+3p+3q+3q^2+q^3$

Thus, the expression is equivalent to:
$E=(p-1{)}^3+(q+1{)}^3$

Substitute $p=6$ and $q=-3$ into this simplified form:
$E=(6-1{)}^3+(-3+1{)}^3$
$E=5^3+(-2{)}^3$
$E=125-8$
$E=117$

Both methods yield the final result of 117.