Question

How do you factor ${{\left( x+y \right)}^{3}}+{{\left( x-y \right)}^{3}}$?

Answer 1

Hint: We need to simplify the cubic polynomials of sum and difference of two numbers. We already have the identity of ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$. For the cube of difference of two numbers, we take ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$. We take the addition of those two identities to find the factored form.

Complete step by step answer:
We need to find the simplified form of ${{\left( a+b \right)}^{3}}$. This is the cube of the sum of two numbers.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We need to multiply the term $\left( a+b \right)$ on both side of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
On the left side of the equation, we get ${{\left( a+b \right)}^{2}}\left( a+b \right)={{\left( a+b \right)}^{3}}$.
On the right side we have $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right)$. We use multiplication and get
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right) \\
 & ={{a}^{2}}.a+a.{{b}^{2}}+2ab\times a+{{a}^{2}}.b+{{b}^{2}}.b+2ab.b \\
 & ={{a}^{3}}+a{{b}^{2}}+2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}+2a{{b}^{2}} \\
 & ={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\
\end{align}$
We also can take another form where
${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Therefore, the simplified form of ${{\left( a+b \right)}^{3}}$ is ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Now we try to find a simplified form of ${{\left( a-b \right)}^{3}}$. Instead of $b$ we take $-b$ and get
\[{{\left( a-b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}\left( -b \right)+3a{{\left( -b \right)}^{2}}+{{\left( -b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}\].
Therefore, ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$.
We take the addition of the two identities to find the solution for ${{\left( a+b \right)}^{3}}+{{\left( a-b \right)}^{3}}$.
$\begin{align}
  & {{\left( a+b \right)}^{3}}+{{\left( a-b \right)}^{3}} \\
 & =\left( {{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \right)+\left( {{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \right) \\
 & =2{{a}^{3}}+6a{{b}^{2}} \\
 & =2a\left( {{a}^{2}}+3{{b}^{2}} \right) \\
\end{align}$
The multiplied form can’t be broken anymore. Therefore, this is the factored form also.
Therefore, \[{{\left( a+b \right)}^{3}}+{{\left( a-b \right)}^{3}}=2a\left( {{a}^{2}}+3{{b}^{2}} \right)\]. We replace values for $a=x;b=y$ and get
\[{{\left( x+y \right)}^{3}}+{{\left( x-y \right)}^{3}}=2x\left( {{x}^{2}}+3{{y}^{2}} \right)\]
Now we take an example to verify the result. We take $x=2;y=1$.
The left-hand side of the equation becomes \[{{\left( x+y \right)}^{3}}+{{\left( x-y \right)}^{3}}={{\left( 2+1 \right)}^{3}}+{{\left( 2-1 \right)}^{3}}={{3}^{3}}+{{1}^{3}}=28\]
The right-hand side of the equation becomes \[2x\left( {{x}^{2}}+3{{y}^{2}} \right)=2\times 2\left( {{2}^{2}}+3\times {{1}^{2}} \right)=4\times 7=28\].
Thus, verified that \[{{\left( x+y \right)}^{3}}+{{\left( x-y \right)}^{3}}=2x\left( {{x}^{2}}+3{{y}^{2}} \right)\].

Note: We also can use the binomial theorem to find the general form and then put the value of 3. We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of the sum of two numbers. So, we put $n=3$.
${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. In this way we also simplify the term of ${{\left( a+b \right)}^{3}}$.