/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Show that $$ (a+b)^{3}=a^{3}... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that $$ (a+b)^{3}=a^{3}+b^{3} $$ if and only if \(a=0\) or \(b=0\) or \(a=-b\).

Short Answer

Expert verified
The given equation \((a+b)^3=a^3+b^3\) holds true if and only if one of the following conditions is met: \(a = 0\), \(b = 0\), or \(a = -b\).

Step by step solution

01

Expand the Left Side using the Binomial Theorem

We start by expanding \((a+b)^3\) using the binomial theorem, which states that for any non-negative integer n and any real numbers a and b: \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \) Applying this to our case with \(n=3\), we have: \[ (a+b)^3 = \sum_{k=0}^{3} \binom{3}{k} a^{3-k} b^k \] Now, we can compute the individual terms of the sum.
02

Compute the Individual Terms of the Sum

We compute the individual terms of the sum as follows: \( \binom{3}{0} a^{3-0} b^0 = 1 \cdot a^3 \cdot 1 = a^3 \\ \binom{3}{1} a^{3-1} b^1 = 3 \cdot a^2 \cdot b = 3a^2b \\ \binom{3}{2} a^{3-2} b^2 = 3 \cdot a \cdot b^2 = 3ab^2 \\ \binom{3}{3} a^{3-3} b^3 = 1 \cdot 1 \cdot b^3 = b^3 \) So, we can rewrite the left side of the equation as: \[ (a+b)^3=a^3+3a^2b+3ab^2+b^3 \]
03

Comparing both sides of the equation and finding the conditions

Now, we have the expanded form of the equation: \[ a^3+3a^2b+3ab^2+b^3 = a^3+b^3 \] By comparing both sides, we observe that the equation holds true when the following term is equal to 0: \[ 3a^2b+3ab^2=0 \] We can factor out \(3ab\) and rewrite the equation as: \[ 3ab(a+b)=0 \] For this equation to be true, one of the factors must equal 0. Therefore, we have three cases: 1) \(a = 0\) 2) \(b = 0\) 3) \(a + b = 0\) which implies \(a = -b\) So, the given equation holds true if and only if \(a = 0\) or \(b = 0\) or \(a = -b\).

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