Question

How do you factor \[{{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}\]?

Answer 1

Hint: From the given question we have to find the factor of \[{{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}\]. By observing it is in the form of \[{{x}^{3}}-{{y}^{3}}\]. We know that formula is \[{{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\] this formula is for difference of cubes as in the question there are perfect cubes, here we have to substitute \[x=a+b\] and \[y=a-b\] by solving we will get the required factors.

 Complete step by step answer:
From the question given \[{{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}\].
Since both the terms are perfect cubes, we will find the factor by using the difference of cubes formula.
We know that the formula for difference of cubes \[{{x}^{3}}-{{y}^{3}}\] is
\[\Rightarrow {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
Here in the above formula, we have to substitute x as \[a+b\] and in place of y as \[a-b\] .
\[\Rightarrow x=a+b\]
\[\Rightarrow y=a-b\]
By substituting in the above formula, we will get,
 \[\Rightarrow {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)\]
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( a+b-\left( a-b \right) \right)\left( {{\left( a+b \right)}^{2}}+\left( a+b \right)\left( a-b \right)+{{\left( a-b \right)}^{2}} \right)\]
In right hand side multiply – with a-b we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( a+b-a+b \right)\left( {{\left( a+b \right)}^{2}}+\left( a+b \right)\left( a-b \right)+{{\left( a-b \right)}^{2}} \right)\]
Now, cancel both a term then we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( b+b \right)\left( {{\left( a+b \right)}^{2}}+\left( a+b \right)\left( a-b \right)+{{\left( a-b \right)}^{2}} \right)\]
Now, add b term then we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( {{\left( a+b \right)}^{2}}+\left( a+b \right)\left( a-b \right)+{{\left( a-b \right)}^{2}} \right)\]
Now, expand \[{{\left( a+b \right)}^{2}}\] as \[{{a}^{2}}+{{b}^{2}}+2ab\] ,
By expanding as above we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( {{a}^{2}}+{{b}^{2}}+2ab+\left( a+b \right)\left( a-b \right)+{{\left( a-b \right)}^{2}} \right)\]
Now, multiply \[\left( a+b \right)\left( a-b \right)\],
by multiplying we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( {{a}^{2}}+{{b}^{2}}+2ab+{{a}^{2}}-{{b}^{2}}+{{\left( a-b \right)}^{2}} \right)\]
Now, expand \[{{\left( a-b \right)}^{2}}\]as \[{{a}^{2}}+{{b}^{2}}-2ab\]
By expanding as above we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( {{a}^{2}}+{{b}^{2}}+2ab+{{a}^{2}}-{{b}^{2}}+{{a}^{2}}+{{b}^{2}}-2ab \right)\]
Now, cancel the like terms with opposite signs,
By eliminating the like terms with opposite signs, we will get,
 \[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( {{a}^{2}}+{{b}^{2}}+{{a}^{2}}+{{a}^{2}} \right)\]
Now, add the like terms,
By adding like terms, we will get,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( 3{{a}^{2}}+{{b}^{2}} \right)\]
Therefore, required factors are \[\left( 2b \right)\], \[\left( 3{{a}^{2}}+{{b}^{2}} \right)\].

 Note:
Students can do the above question by expanding the \[{{\left( a+b \right)}^{3}}\] and \[{{\left( a-b \right)}^{3}}\]i.e.,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}-\left( {{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} \right)\]
by simplifying the above equations, we will get the,
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( 3{{a}^{2}}+{{b}^{2}} \right)\]
If students are asked to find the complex factors then \[\left( 3{{a}^{2}}+{{b}^{2}} \right)\] can be divided into \[\sqrt{3}a+ib\], \[\sqrt{3}a-ib\]
Therefore, the factors will be
\[\Rightarrow {{\left( a+b \right)}^{3}}-{{\left( a-b \right)}^{3}}=\left( 2b \right)\left( \sqrt{3}a+ib \right)\left( \sqrt{3}a-ib \right)\]