Math Problem Statement

Find the following product: (3y + 3)^3

Solution

To solve the problem (3y+3)3(3y + 3)^3, we need to expand the cubic binomial expression using the binomial theorem or by multiplying step by step.

Step 1: Recall the Binomial Theorem

The binomial theorem states: (a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

For (3y+3)3(3y + 3)^3, we set a=3ya = 3y, b=3b = 3, and n=3n = 3.

Step 2: Expand (3y+3)3(3y + 3)^3

Using the binomial theorem, the expansion is: (3y+3)3=(30)(3y)3(3)0+(31)(3y)2(3)1+(32)(3y)1(3)2+(33)(3y)0(3)3(3y + 3)^3 = \binom{3}{0}(3y)^3(3)^0 + \binom{3}{1}(3y)^2(3)^1 + \binom{3}{2}(3y)^1(3)^2 + \binom{3}{3}(3y)^0(3)^3

Step 3: Simplify Each Term

  1. First term: (30)(3y)3(3)0=1(3y)31=27y3\binom{3}{0}(3y)^3(3)^0 = 1 \cdot (3y)^3 \cdot 1 = 27y^3

  2. Second term: (31)(3y)2(3)1=3(3y)23=39y23=81y2\binom{3}{1}(3y)^2(3)^1 = 3 \cdot (3y)^2 \cdot 3 = 3 \cdot 9y^2 \cdot 3 = 81y^2

  3. Third term: (32)(3y)1(3)2=3(3y)9=327y=81y\binom{3}{2}(3y)^1(3)^2 = 3 \cdot (3y) \cdot 9 = 3 \cdot 27y = 81y

  4. Fourth term: (33)(3y)0(3)3=1127=27\binom{3}{3}(3y)^0(3)^3 = 1 \cdot 1 \cdot 27 = 27

Step 4: Combine the Results

Now, combine all the terms: (3y+3)3=27y3+81y2+81y+27(3y + 3)^3 = 27y^3 + 81y^2 + 81y + 27

Final Answer:

(3y+3)3=27y3+81y2+81y+27(3y + 3)^3 = 27y^3 + 81y^2 + 81y + 27


Would you like additional steps explained or clarification?

Related Questions:

  1. How do you expand a polynomial like (x+y)n(x + y)^n for larger values of nn?
  2. What is Pascal's triangle, and how is it related to binomial expansion?
  3. How do you multiply binomials step-by-step without using the binomial theorem?
  4. Can the binomial theorem be applied to negative exponents or fractional exponents?
  5. How can we factor 27y3+81y2+81y+2727y^3 + 81y^2 + 81y + 27 back into (3y+3)3(3y + 3)^3?

Tip:

For higher powers, Pascal’s triangle is a quick way to determine binomial coefficients.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Multiplication
Binomial Theorem

Formulas

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k

Theorems

Binomial Theorem

Suitable Grade Level

Grades 9-12