Math Problem Statement

(x-3y)^3

Solution

To expand the expression (x3y)3(x - 3y)^3, we can use the binomial theorem, which states:

(a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

In this case, a=xa = x, b=3yb = -3y, and n=3n = 3. Let's apply the binomial expansion:

(x3y)3=k=03(3k)x3k(3y)k(x - 3y)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} (-3y)^k

Step 1: Calculate the Binomial Coefficients

(30)=1,(31)=3,(32)=3,(33)=1\binom{3}{0} = 1, \quad \binom{3}{1} = 3, \quad \binom{3}{2} = 3, \quad \binom{3}{3} = 1

Step 2: Expand the Expression

Now, we expand each term:

  • When k=0k = 0: (30)x30(3y)0=1x31=x3\binom{3}{0} x^{3-0} (-3y)^0 = 1 \cdot x^3 \cdot 1 = x^3

  • When k=1k = 1: (31)x31(3y)1=3x2(3y)=9x2y\binom{3}{1} x^{3-1} (-3y)^1 = 3 \cdot x^2 \cdot (-3y) = -9x^2y

  • When k=2k = 2: (32)x32(3y)2=3x9y2=27xy2\binom{3}{2} x^{3-2} (-3y)^2 = 3 \cdot x \cdot 9y^2 = 27xy^2

  • When k=3k = 3: (33)x33(3y)3=11(27y3)=27y3\binom{3}{3} x^{3-3} (-3y)^3 = 1 \cdot 1 \cdot (-27y^3) = -27y^3

Step 3: Combine All Terms

Finally, add all the terms together:

(x3y)3=x39x2y+27xy227y3(x - 3y)^3 = x^3 - 9x^2y + 27xy^2 - 27y^3

This is the expanded form of (x3y)3(x - 3y)^3.

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Relative Questions

  1. How does the binomial theorem help in expanding binomials raised to higher powers?
  2. What are binomial coefficients, and how are they calculated?
  3. Can you apply the binomial expansion to (x+y)4(x + y)^4?
  4. How can you use Pascal's triangle to find binomial coefficients?
  5. What are the general properties of binomials raised to different powers?

Tip

When expanding binomials using the binomial theorem, make sure to keep track of the signs carefully, especially when dealing with negative terms.

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Math Problem Analysis

Mathematical Concepts

Binomial theorem
Binomial coefficients
Polynomial expansion

Formulas

Binomial theorem formula (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Theorems

Binomial theorem

Suitable Grade Level

Grades 11-12