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Chapter 2: Problem 46

Verify that \((x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\).

Short Answer

Expert verified
Using the binomial theorem formula, we expanded and simplified \((x+y)^3\) as follows: \((x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3\). This is the same as the given equation, thus verifying that \((x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\).

Step by step solution

01

Identify the variables a, b, and n in the given problem

In this case, the values for a, b, and n are: \(a = x\) \(b = y\) \(n = 3\)
02

Apply the binomial theorem formula

Using the given values of a, b, and n, we can apply the binomial theorem formula: \((x + y)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} y^k\)
03

Calculate the binomial coefficients

Calculate the binomial coefficients for each term in the sum: \(\binom{3}{0} = \frac{3!}{0!(3-0)!} = 1\) \(\binom{3}{1} = \frac{3!}{1!(3-1)!} = 3\) \(\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3\) \(\binom{3}{3} = \frac{3!}{3!(3-3)!} = 1\)
04

Substitute the binomial coefficients into the sum

Now we can substitute the binomial coefficients into the sum: \((x + y)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} y^k = 1 \cdot x^3 \cdot y^0 + 3 \cdot x^2 \cdot y^1 + 3 \cdot x^1 \cdot y^2 + 1 \cdot x^0 \cdot y^3\)
05

Simplify the expression

Simplify the expression by applying the rules for exponents: \((x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3\)
06

Compare the expanded expression to the given one

Compare to the right-hand side of the original equation: \(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\) Since the expanded expression and the given equation are the same, we have verified that: \((x + y)^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}\)

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