Factorise $ {a^6} - {b^6} $

Hint: Use the formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ for factorisation. The formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ states that the difference between the squares of the two numbers is equal to the product of the sum and difference of those two numbers.

Complete step-by-step answer:
The given expression is $ {a^6} - {b^6} $ .
We know that, $ {a^{mn}} = {\left( {{a^m}} \right)^n} $
6 can be written as $ 2 \cdot 3 $ .
 $ {a^6} $ can be written as $ {a^{2 \cdot 3}} = {\left( {{a^2}} \right)^3} $ . Also, $ {b^6} $ can be written as $ {b^{2 \cdot 3}} = {\left( {{b^2}} \right)^3} $
The expression $ {a^6} - {b^6} $ can be written as $ {\left( {{a^2}} \right)^3} - {\left( {{b^2}} \right)^3} $
We can use the formula, $ {m^3} - {n^3} = \left( {m - n} \right)\left( {{m^2} + mn + {n^2}} \right) $ to factorise the given equation.
 $ {a^6} - {b^6} = {\left( {{a^2}} \right)^3} - {\left( {{b^2}} \right)^3} $
After comparing $ {\left( {{a^2}} \right)^3} - {\left( {{b^2}} \right)^3} $ with $ {m^3} - {n^3} $ , we get, $ m = {a^2},{\rm{ }}n = {b^2} $ .
Substitute the values of $ m = {a^2},{\rm{ }}n = {b^2} $ in the equation $ {m^3} - {n^3} = \left( {m - n} \right)\left( {{m^2} + mn + {n^2}} \right) $ .
\[
\Rightarrow {a^6} - {b^6} = {\left( {{a^2}} \right)^3} - {\left( {{b^2}} \right)^3}\\
 = \left( {{a^2} - {b^2}} \right)\left( {{{\left( {{a^2}} \right)}^2} + {a^2}{b^2} + {{\left( {{b^2}} \right)}^2}} \right)
\]
We can use the formula $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ to simplify the equation.
 $
\Rightarrow {a^6} - {b^6} = \left( {{a^2} - {b^2}} \right)\left( {{{\left( {{a^2}} \right)}^2} + {a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right)\\
 = \left( {a + b} \right)\left( {a - b} \right)\left( {{{\left( {{a^2}} \right)}^2} + {a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right)
 $
To further factorise the equation, we can add and subtract $ {a^2}{b^2} $ to the expression $ \Rightarrow \left( {{{\left( {{a^2}} \right)}^2} + {a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right) $ .
 \[
\Rightarrow {a^6} - {b^6} = \left( {a + b} \right)\left( {a - b} \right)\left( {{{\left( {{a^2}} \right)}^2} + {a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right)\\
 = \left( {a + b} \right)\left( {a - b} \right)\left( {{{\left( {{a^2}} \right)}^2} + {a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2} + {a^2} \cdot {b^2} - {a^2} \cdot {b^2}} \right)\\
 = \left( {a + b} \right)\left( {a - b} \right)\left( {\left( {{{\left( {{a^2}} \right)}^2} + 2{a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right) - {a^2} \cdot {b^2}} \right)
\]
\[\left( {{{\left( {{a^2}} \right)}^2} + 2{a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right)\] is the term which can be compared with $ {m^2} + 2mn + {n^2} $ .
We get, $ m = {a^2},{\rm{ }}n = {b^2} $ and substitute them in the equation $ {\left( {m + n} \right)^2} = {m^2} + 2mn + {n^2} $ .
 $ {\left( {{a^2} + {b^2}} \right)^2} = {\left( {{a^2}} \right)^2} + 2{a^2}{b^2} + {\left( {{b^2}} \right)^2} $
So,
\[
\Rightarrow {a^6} - {b^6} = \left( {a + b} \right)\left( {a - b} \right)\left( {\left( {{{\left( {{a^2}} \right)}^2} + 2{a^2} \cdot {b^2} + {{\left( {{b^2}} \right)}^2}} \right) - {a^2} \cdot {b^2}} \right)\\
 = \left( {a + b} \right)\left( {a - b} \right)\left( {{{\left( {{a^2} + {b^2}} \right)}^2} - {a^2} \cdot {b^2}} \right)\\
 = \left( {a + b} \right)\left( {a - b} \right)\left( {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} \right)
\]
Use the formula $ {m^2} - {n^2} = \left( {m + n} \right)\left( {m - n} \right) $ .
Now we should compare the expression \[\left( {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} \right)\] with $ {m^2} - {n^2} $ .
We get, $ m = \left( {{a^2} + {b^2}} \right),{\rm{ }}n = ab $ .
After substituting $ m = \left( {{a^2} + {b^2}} \right),{\rm{ }}n = ab $ in the equation $ {m^2} - {n^2} = \left( {m + n} \right)\left( {m - n} \right) $ , we get,
\[\Rightarrow {\left( {{a^2} + {b^2}} \right)^2} - {\left( {ab} \right)^2} = \left( {{a^2} + {b^2} + ab} \right)\left( {{a^2} + {b^2} - ab} \right)\]
Therefore, \[{a^6} - {b^6} = \left( {a + b} \right)\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\left( {{a^2} + {b^2} - ab} \right)\] is the right answer.

Additional information: Factorisation of polynomials is done by separating it in the product of factors which cannot be reduced further. In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials.
There are some of the important identities which are mostly used to either factorise of an expression.
 $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
The square of the sum of two numbers can be given as the sum of the squares of the individual numbers added to twice the product of those two numbers.
 $ {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab $
The square of the difference of two numbers can be given by subtracting twice the product of the numbers from the sum of the squares of the individual numbers.
 $ {a^2} - {b^2} = \left( {a + b} \right) \cdot \left( {a - b} \right) $
The difference between the squares of the two numbers is equal to the product of the sum and difference of those two numbers.


Note: Please take special care while considering the symbols of different operations i.e. addition, subtraction, multiplication. Students should follow the BODMAS order of operation which denotes, solving first the brackets, then division, multiplication, addition and at last subtraction.