Expand (2a +36) (2a-3b

Use the binomial expansion theorem to find each term. The binomial theorem states
(
a
+
b
)
n
=
n
∑
k
=
0

n
C
k
⋅
(
a
n
−
k
b
k
)
(
a
+
b
)
n
=
∑
k
=
0
n
⁡
n
C
k
⋅
(
a
n
-
k
b
k
)
.
3
∑
k
=
0

3
!
(
3
−
k
)
!
k
!
⋅
(
2
a
)
3
−
k
⋅
(
−
3
b
)
k
∑
k
=
0
3
⁡
3
!
(
3
-
k
)
!
k
!
⋅
(
2
a
)
3
-
k
⋅
(
-
3
b
)
k
Expand the summation.
3
!
(
3
−
0
)
!
0
!
⋅
(
2
a
)
3
−
0
⋅
(
−
3
b
)
0
+
3
!
(
3
−
1
)
!
1
!
⋅
(
2
a
)
3
−
1
⋅
(
−
3
b
)
+
3
!
(
3
−
2
)
!
2
!
⋅
(
2
a
)
3
−
2
⋅
(
−
3
b
)
2
+
3
!
(
3
−
3
)
!
3
!
⋅
(
2
a
)
3
−
3
⋅
(
−
3
b
)
3
3
!
(
3
-
0
)
!
0
!
⋅
(
2
a
)
3
-
0
⋅
(
-
3
b
)
0
+
3
!
(
3
-
1
)
!
1
!
⋅
(
2
a
)
3
-
1
⋅
(
-
3
b
)
+
3
!
(
3
-
2
)
!
2
!
⋅
(
2
a
)
3
-
2
⋅
(
-
3
b
)
2
+
3
!
(
3
-
3
)
!
3
!
⋅
(
2
a
)
3
-
3
⋅
(
-
3
b
)
3
Simplify the exponents for each term of the expansion.
1
⋅
(
2
a
)
3
⋅
(
−
3
b
)
0
+
3
⋅
(
2
a
)
2
⋅
(
−
3
b
)
+
3
⋅
(
2
a
)
⋅
(
−
3
b
)
2
+
1
⋅
(
2
a
)
0
⋅
(
−
3
b
)
3
1
⋅
(
2
a
)
3
⋅
(
-
3
b
)
0
+
3
⋅
(
2
a
)
2
⋅
(
-
3
b
)
+
3
⋅
(
2
a
)
⋅
(
-
3
b
)
2
+
1
⋅
(
2
a
)
0
⋅
(
-
3
b
)
3
Simplify each term.
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8
a
3
−
36
a
2
b
+
54
a
b
2
−
27
b
3
8
a
3
-
36
a
2
b
+
54
a
b
2
-
27
b
3

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4y - 6 + y^2 - 9 = y^2 + 4y - 15y^2 term terms- termstotal of 3 terms when simplified