Factorise ${{a}^{7}}-a{{b}^{6}}$.

Hint:Factorisation is a method of expressing or writing a given number in the form of the product of other numbers. The numbers present in the product are called as factors of the original number.

Formula used:
${{x}^{3}}-{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})$
${{x}^{3}}+{{y}^{3}}=(x-y)({{x}^{2}}+xy+{{y}^{2}})$
${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$, where x and y are real numbers.

Complete step by step answer:
Let us first understand what is meant by the term factorisation. Factorisation is a method of expressing or writing a given number in the form of the product of other numbers. The numbers present in the product are called as factors of the original number. For example, the number 6 can be written in the form of the product of the numbers 2 and 3 i.e. $6=3\times 2$. Here, 3 and 2 are the factors of 6.

The method of factorisation can also be used for expressions containing variables. Here, the question is that we have to factorise the expression ${{a}^{7}}-a{{b}^{6}}$.Let this expression be expression (i).Let us analyse the expression (i) and try to simplify it.We can see that the expression has two terms. It is actually a subtraction of two terms. We can also see that the two terms are multiples of ‘a’. This means that the two terms have ‘a’ as a common factor.Therefore, let us multiply and divide the expression by ‘a’.

We get that
${{a}^{7}}-a{{b}^{6}}=\dfrac{a}{a}\times \left( {{a}^{7}}-a{{b}^{6}} \right)$
$\Rightarrow {{a}^{7}}-a{{b}^{6}}=a\times \left( \dfrac{{{a}^{7}}}{a}-\dfrac{a{{b}^{6}}}{a} \right)$
$\Rightarrow {{a}^{7}}-a{{b}^{6}}=a\left( {{a}^{6}}-{{b}^{6}} \right)$ ….. (ii)
The term $\left( {{a}^{6}}-{{b}^{6}} \right)={{\left( {{a}^{2}} \right)}^{3}}-{{\left( {{b}^{2}} \right)}^{3}}$
Now, we can use the identity ${{x}^{3}}-{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})$

With this we get $\left( {{a}^{6}}-{{b}^{6}} \right)=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{\left( {{a}^{2}} \right)}^{2}}-{{a}^{2}}{{b}^{2}}+{{\left( {{b}^{2}} \right)}^{2}} \right)$
$\Rightarrow \left( {{a}^{6}}-{{b}^{6}} \right)=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{4}}-{{a}^{2}}{{b}^{2}}+{{b}^{4}} \right)$
Substitute this value in (ii).
$\therefore{{a}^{7}}-a{{b}^{6}}=a\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{4}}-{{a}^{2}}{{b}^{2}}+{{b}^{4}} \right)$

Hence, the given expression is factored.

Note:We can also the write the term $\left( {{a}^{6}}-{{b}^{6}} \right)$ as $\left( {{a}^{6}}-{{b}^{6}} \right)={{\left( {{a}^{3}} \right)}^{2}}-{{\left( {{b}^{3}} \right)}^{2}}$.
Then we can use the identity that says ${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$.
Then,
$\Rightarrow \left( {{a}^{6}}-{{b}^{6}} \right)=\left( {{a}^{3}}+{{b}^{3}} \right)\left( {{a}^{3}}-{{b}^{3}} \right)$
Next, by using the identities ${{x}^{3}}+{{y}^{3}}=(x-y)({{x}^{2}}+xy+{{y}^{2}})$ and ${{x}^{3}}-{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})$ we get that
$\Rightarrow \left( {{a}^{6}}-{{b}^{6}} \right)=(a-b)({{a}^{2}}+ab+{{b}^{2}})(a+b)({{a}^{2}}-ab+{{b}^{2}})$
With this we get that ${{a}^{7}}-a{{b}^{6}}=a(a-b)({{a}^{2}}+ab+{{b}^{2}})(a+b)({{a}^{2}}-ab+{{b}^{2}})$