Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$

Question

Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$

Why does this work? I am confused as to why this happens.

Answer 1

Since pictures don't appear in the comments well,

Answer 2

Let's break the algebra down, one step at a time. Trying to use colors to keep track of what comes from what.

You have that $$(a+b)^3 = \color{green}{(a+b)^2}\cdot (\color{#AA33FF}{a}+\color{#11AAFF}{b})$$ I assume you know $(a+b)^2 = a^2+2ab+b^2$, from which $$ (a+b)^3 = \color{green}{(a^2+2ab+b^2)}\cdot (\color{#AA33FF}{a}+\color{#11AAFF}{b}) $$ Now, distributing (i.e., $c(a+b)=ca+cb$), $$ (a+b)^3 = \color{green}{(a^2+2ab+b^2)}\cdot \color{#AA33FF}{a}+\color{green}{(a^2+2ab+b^2)}\cdot \color{#11AAFF}{b} $$ Distributing again, $$ (a+b)^3 = a^2\cdot \color{#AA33FF}{a}+2ab\cdot \color{#AA33FF}{a}+b^2\cdot \color{#AA33FF}{a} + a^2\cdot \color{#11AAFF}{b}+2ab\cdot \color{#11AAFF}{b}+b^2\cdot \color{#11AAFF}{b} $$ that is $$ (a+b)^3 = a^3+\color{red}{ 2a^2b}+\color{blue}{ a b^2} + \color{red} {a^2b}+\color{blue}{ 2ab^2}+b^3 $$ and regrouping the blue and red terms together, $$ (a+b)^3 = a^3+\color{red}{3a^2b}+\color{blue}{3ab^2}+b^3 $$ giving the result.

Answer 3

In overly pedantic detail: \begin{align} (a+b)^3 &= (a+b)[(a+b)(a+b)] \tag{by definition} \\ &= (a+b)[a(a+b) + b(a+b)] \tag{distribution} \\ &= (a+b)[a^2 + ab + ba + b^2] \tag{distribution} \\ &= (a+b)[a^2 + 2ab + b^2] \tag{commutativity} \\ &= a[a^2 + 2ab + b^2] + b[a^2 + 2ab + b^2] \tag{distribution} \\ &= a^3 + a2ab + ab^2 + ba^2 + b2ab + b^3 \tag{distribution} \\ &= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 \tag{commutativity} \\ &= a^3 + 3a^2b + 3ab^2 + b^3. \tag{collect like terms} \end{align}

Answer 4

If you want something step by step use the distributive property and associative property: $$ \begin{array}{rll} (a+b)^{3}= & \underbrace{(\color{red}{a}+\color{red}{b})}_{=x} \cdot \underbrace{(\color{green}{a}+\color{green}{b})}_{=y} \cdot (\color{blue}{a}+\color{blue}{b}) &\\ =&\underbrace{\color{red}{x} \cdot \color{green}{y}}_{=z} \cdot (\color{blue}{a}+\color{blue}{b}) &\\ =& z\cdot (\color{blue}{a}+\color{blue}{b}) &\\ =& z\cdot \color{blue}{a}+z\cdot\color{blue}{b} & \mbox{distributive property} \\ =& (x\cdot y)\cdot \color{blue}{a}+(x\cdot y)\cdot\color{blue}{b} & z=(x\cdot y) \\ =& x\cdot (y\cdot \color{blue}{a})+x\cdot (y\cdot\color{blue}{b}) & \mbox{associative property} \\ =& x\cdot \big((\color{green}{a}+\color{green}{b})\cdot \color{blue}{a}\big) + x\cdot \big((\color{green}{a}+\color{green}{b})\cdot\color{blue}{b}\big) & y=\color{green}{a}+\color{green}{b} \\ =& \underbrace{x\cdot \big(\color{green}{a}\color{blue}{a}+\color{green}{b}\color{blue}{a}\big)}_{} + \underbrace{x\cdot \big(\color{green}{a}\color{blue}{b}+\color{green}{b}\color{blue}{b}\big)}_{} & \mbox{distributive property} \\ =& x\cdot(\color{green}{a}\color{blue}{a})+x\cdot(\color{green}{b}\color{blue}{a}) + x\cdot(\color{green}{a}\color{blue}{b})+x\cdot(\color{green}{b}\color{blue}{b}) & \mbox{distributive property} \\ =& x\color{green}{a}\color{blue}{a}\;+ \\ &\hspace{6mm} + 2x\color{green}{a}\color{blue}{b}\;+ \\ &\hspace{18mm} + x\color{green}{b}\color{blue}{b} \\ =& (\color{red}{a}+\color{red}{b})\color{green}{a}\color{blue}{a}\;+ \\ &\hspace{5mm} + 2(\color{red}{a}+\color{red}{b})\color{green}{a}\color{blue}{b}\;+ \\ &\hspace{15mm} + (\color{red}{a}+\color{red}{b})\color{green}{b}\color{blue}{b} \\ =& \color{red}{a}\color{green}{a}\color{blue}{a}+\color{red}{b}\color{green}{a}\color{blue}{a}\;+ &\mbox{distributive property}\\ &\hspace{5mm} + 2(\color{red}{a}\color{green}{a}\color{blue}{b}+\color{red}{b}\color{green}{a}\color{blue}{b})\;+ & \mbox{distributive property}\\ &\hspace{15mm} + \color{red}{a}\color{green}{b}\color{blue}{b}+\color{red}{b}\color{green}{b}\color{blue}{b} &\mbox{distributive property} \\ =& \color{red}{a}\color{green}{a}\color{blue}{a}+\color{red}{b}\color{green}{a}\color{blue}{a}\;+ &\\ &\hspace{8mm} + 2\color{red}{a}\color{green}{a}\color{blue}{b}+2\color{red}{b}\color{green}{a}\color{blue}{b}\;+ & \mbox{distributive property}\\ &\hspace{21mm} + \color{red}{a}\color{green}{b}\color{blue}{b}+\color{red}{b}\color{green}{b}\color{blue}{b} \\ =& a^3+a^2b+ &\\ &\hspace{8mm} + 2{a}^2{b}+2{a}{b}^2\;+ & \\ &\hspace{21mm} + {a}{b}^2+{b}^3 \\ =& a^3+3a^2b+3ab^2+b^3 \end{array} $$

Answer 5

Algebra meets combinatorics. The exponent $3$ tells us to expand the product of three binomials \begin{align*} (a+b)^3=(a+b)(a+b)(a+b) \end{align*} From each of the three factors $a+b$ we have to select either $a$ or $b$.

Let's list the different possibilities:

• From each binomial we select an $a$. Since there is only one way to do so we obtain \begin{align*} aaa\qquad\rightarrow\qquad \color{blue}{1a^3}\qquad\qquad\ \ \end{align*}
• Number of ways to select two $a$'s and one $b$ \begin{align*} aab,\ aba,\ baa\qquad\rightarrow\qquad aab+aba+baa=\color{blue}{3a^2b} \end{align*}
• Number of ways to select two $b$'s and one $a$ \begin{align*} bba,\ bab,\ abb\qquad\rightarrow\qquad bba+bab+abb=\color{blue}{3ab^2} \end{align*}
• Number of ways to select three $b$'s \begin{align*} bbb\qquad\rightarrow\qquad \color{blue}{1b^3}\qquad\qquad\ \ \end{align*}
• No other possibilities.

We conclude \begin{align*} (a+b)^3=a^3+3a^2b+3ab^2+b^3 \end{align*}

Hint: Observe the nice symmetry between $a$ and $b$.

Challenge: Try a similar approach with $(a+b)^\color{blue}{4}$ and verify the result algebraically. This is more complex but still feasible .