Question

Find what to multiply together to get an expression ${{a}^{7}}-a{{b}^{6}}$ . OR Factorise .

Answer 1

Hint: Factorising is the reverse of calculating the product of factors. In order to factorise a quadratic, you need to find the factors which, when multiplied together, equal the original quadratic. [Use formulae $\left( {{a}^{2}}-{{b}^{2}} \right),\left( {{a}^{3}}-{{b}^{3}} \right)$ and$\left( {{a}^{3}}+{{b}^{3}} \right)$]

Complete step-by-step answer:
In the given expression, a is a common factor of the first and second terms and then the given expression ${{a}^{7}}-a{{b}^{6}}$ can be factorised as follows
 ${{a}^{7}}-a{{b}^{6}}=a\left( {{a}^{6}}-{{b}^{6}} \right)$
Rearranging the terms by using indices formula, we get
${{a}^{7}}-a{{b}^{6}}=a\left[ {{\left( {{a}^{3}} \right)}^{2}}-{{\left( {{b}^{3}} \right)}^{2}} \right]$
So ${{x}^{2}}-{{y}^{2}}$ can be written in factorised form as $\left( x+y \right)\left( x-y \right)$
This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down the factors. These types of quadratics are very simple to factorise.
${{a}^{7}}-a{{b}^{6}}=a\left( {{a}^{3}}+{{b}^{3}} \right)\left( {{a}^{3}}-{{b}^{3}} \right)$
We are working with the sum of two cubes. We know that ${{x}^{3}}+{{y}^{3}}=(x+y)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$, so we need to identify x and y.
We can replace x and y in the factorised form of the expression for the sum of two cubes with a and b.
${{a}^{7}}-a{{b}^{6}}=a(a+b)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\left( {{a}^{3}}-{{b}^{3}} \right)$
Now we are working with the difference of two cubes. We know that${{x}^{3}}-{{y}^{3}}=(x-y)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$, so we need to identify x and y.
Again, we can replace x and y in the factorised form of the expression for the difference of two cubes with a and b.
${{a}^{7}}-a{{b}^{6}}=a(a+b)\left({{a}^{2}}-ab+{{b}^{2}}\right)(a-b)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
Rearranging the terms, we get
${{a}^{7}}-a{{b}^{6}}=a(a+b)(a-b)\left({{a}^{2}}-ab+{{b}^{2}}\right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
Hence the factors of \[{{a}^{7}}-a{{b}^{6}}\] are $a,(a+b),(a-b),\left( {{a}^{2}}-ab+{{b}^{2}} \right)\text{ and }\left( {{a}^{2}}+ab+{{b}^{2}} \right)$

Note: The possibility for the mistake is that you might get confused with cancel terms that are not common factors of either the numerator or denominator of a quotient of expressions.