Question

How do you factor \[{a^3} - {b^3}\].

Answer 1

Hint: In this question, we will use the concept of the algebraic identity to find the factor of the given expression. Factor is the way to write the expression in the product of two expressions such that it gives the given expression.

Complete step by step answer:
We know that we need to find the factor of \[\left( {{a^3} - {b^3}} \right)\].
First, we will use the below formula for the factor of the given expression as,
\[ \Rightarrow {\left( {a - b} \right)^3} = \left( {a - b} \right)\left( {a - b} \right)\left( {a - b} \right)\]
We take Right Hand Side and calculate as,
\[ \Rightarrow \left( {a - b} \right)\left( {a - b} \right)\left( {a - b} \right)\]
Multiply each of them. First, we multiply two of them and then we multiply the third one.
\[ \Rightarrow \left( {{a^2} - ab - ba + {b^2}} \right)\left( {a - b} \right)\]
After simplification it can be written as,
\[\left( {{a^2} - 2ab + {b^2}} \right)\left( {a - b} \right)\]
Again, multiply by remaining part and we will get,
\[ \Rightarrow \left( {{a^3} - {a^2}b - 2{a^2}b + 2a{b^2} + a{b^2} - {b^3}} \right)\]
By solving the above equation, the result would be as below.
\[ \Rightarrow \left( {{a^3} - {b^3} - 3{a^2}b + 3a{b^2}} \right)\]
Next, take both LHS and RHS side, then
\[ \Rightarrow {\left( {a - b} \right)^3} = \left( {{a^3} - {b^3} - 3{a^2}b + 3a{b^2}} \right)\]
It is written as simply as below.
\[ \Rightarrow {\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\]
Take the\[ - 3ab\left( {a - b} \right)\]part in LHS side then,
\[ \Rightarrow {\left( {a - b} \right)^3} + 3ab\left( {a - b} \right) = {a^3} - {b^3}\]
Next, we take the LHS side for calculating.
\[ \Rightarrow {\left( {a - b} \right)^3} + 3ab\left( {a - b} \right)\]
In the above equation we take the common part outside, and then the new equation is:
\[ \Rightarrow \left( {a - b} \right)\left( {{{\left( {a - b} \right)}^2} + 3ab} \right)\]
We know that the formula of \[{\left( {a - b} \right)^2}\]. Put that in above equation
\[\left( {a - b} \right)\left( {\left( {{a^2} + {b^2} - 2ab} \right) + 3ab} \right)\]
This equation is written in simple form as below.
\[\left( {a - b} \right)\left( {{a^2} + {b^2} - 2ab + 3ab} \right)\]
By the calculating of above equation, the result would be come
\[ \Rightarrow \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
There for the result is:
\[ \Rightarrow \left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]

Therefore, the factor of \[{a^3} - {b^3}\] are\[\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\].

Note:
In this question, we find the factor of \[{a^3} - {b^3}\]. The formula for finding the factor of \[{a^3} - {b^3}\] type is derived from the \[{\left( {a - b} \right)^3}\] expression. If the number which comes in place of a and b is a perfect cube. Then the factor can easily find out.