Use the special product formula to expand the following expression completely: $(2a+3b)^3$
Answer (1)
Identify the special product formula: ( x + y ) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3 .
Substitute x = 2 a and y = 3 b into the formula: ( 2 a + 3 b ) 3 = ( 2 a ) 3 + 3 ( 2 a ) 2 ( 3 b ) + 3 ( 2 a ) ( 3 b ) 2 + ( 3 b ) 3 .
Simplify each term: ( 2 a ) 3 = 8 a 3 , 3 ( 2 a ) 2 ( 3 b ) = 36 a 2 b , 3 ( 2 a ) ( 3 b ) 2 = 54 a b 2 , and ( 3 b ) 3 = 27 b 3 .
Combine the terms to get the final expanded expression: 8 a 3 + 36 a 2 b + 54 a b 2 + 27 b 3 .
Explanation
Understanding the Problem We are asked to expand the expression ( 2 a + 3 b ) 3 using the special product formula.
Identifying the Formula The special product formula that applies here is ( x + y ) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3 .
Applying the Formula Apply the special product formula ( x + y ) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3 with x = 2 a and y = 3 b . Substituting these values into the formula, we get: ( 2 a + 3 b ) 3 = ( 2 a ) 3 + 3 ( 2 a ) 2 ( 3 b ) + 3 ( 2 a ) ( 3 b ) 2 + ( 3 b ) 3
Simplifying Each Term Now, let's simplify each term:
( 2 a ) 3 = 2 3 ⋅ a 3 = 8 a 3
3 ( 2 a ) 2 ( 3 b ) = 3 ( 4 a 2 ) ( 3 b ) = 3 ⋅ 4 ⋅ 3 ⋅ a 2 ⋅ b = 36 a 2 b
3 ( 2 a ) ( 3 b ) 2 = 3 ( 2 a ) ( 9 b 2 ) = 3 ⋅ 2 ⋅ 9 ⋅ a ⋅ b 2 = 54 a b 2
( 3 b ) 3 = 3 3 ⋅ b 3 = 27 b 3
Combining Terms Combining the simplified terms, we get the final expanded expression: ( 2 a + 3 b ) 3 = 8 a 3 + 36 a 2 b + 54 a b 2 + 27 b 3
Examples
Understanding how to expand expressions like ( 2 a + 3 b ) 3 is useful in various fields, such as engineering and physics, where polynomial expansions are frequently used to model physical phenomena. For example, when calculating the volume of a cube with sides of length ( 2 a + 3 b ) , the expanded form helps in understanding how each term contributes to the total volume. Moreover, in financial mathematics, such expansions can be used to model compound interest or growth scenarios where the principal amount grows at a rate proportional to ( 2 a + 3 b ) . This algebraic skill provides a foundation for more complex mathematical modeling and problem-solving.
