Question

How do you factor ${{x}^{3}}+{{y}^{3}}$ ?

Answer 1

Hint: Factorization of ${{x}^{3}}+{{y}^{3}}$is $\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$. In this question, we need to find the factors by applying a particular method. So, we will apply the binomial theorem here. We will apply a binomial theorem with n = 2 and n = 3 where n is the power.

Complete step by step answer:
Now, let’s solve the question.
As we know,
$\Rightarrow {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}.......(i)$
So, can we write ${{\left( x+y \right)}^{3}}={{\left( x+y \right)}^{2}}\left( x+y \right)$
Now, by using equation(i) we can write:
$\Rightarrow {{\left( x+y \right)}^{3}}=\left( {{x}^{2}}+2xy+{{y}^{2}} \right)\left( x+y \right)$
Next step is to open the brackets and multiply each and every term. Let’s see how this will be done.
$\Rightarrow {{\left( x+y \right)}^{3}}={{x}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}}$
Now, group the terms having power as 3.
$\Rightarrow {{\left( x+y \right)}^{3}}=\left( {{x}^{3}}+{{y}^{3}} \right)+3{{x}^{2}}y+3x{{y}^{2}}$
In the next step, we have to move the last two terms to the other side of the equation. We will get:
$\Rightarrow {{\left( x+y \right)}^{3}}-3{{x}^{2}}y-3x{{y}^{2}}=\left( {{x}^{3}}+{{y}^{3}} \right)$
Are there any common terms now? If it’s there then solve the equation further.
$\Rightarrow {{\left( x+y \right)}^{3}}-3xy(x+y)=\left( {{x}^{3}}+{{y}^{3}} \right)$
$\Rightarrow \left( x+y \right)\left[ {{\left( x+y \right)}^{2}}-3xy \right]={{x}^{3}}+{{y}^{3}}$
By using equation(i), solve the above expression. We get:
$\Rightarrow \left( x+y \right)\left[ {{x}^{2}}+2xy+{{y}^{2}}-3xy \right]={{x}^{3}}+{{y}^{3}}$
On solving further, we get:
$\Rightarrow \left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)={{x}^{3}}+{{y}^{3}}$
So, finally we got the factors of ${{x}^{3}}+{{y}^{3}}$.

Note: There is an alternative method of finding the factors. Let us discuss that also. In this method, we have to add the term and subtract the same term in order to find the factors. Let’s see how.
First, write the expression given in question:
$\Rightarrow {{x}^{3}}+{{y}^{3}}$
Now, add the term ${{x}^{2}}y$ and subtract the same term in above expression.
$\Rightarrow {{x}^{3}}+{{x}^{2}}y-{{x}^{2}}y+{{y}^{3}}$
Next step is to extract the common terms by grouping them.
$\Rightarrow {{x}^{2}}(x+y)-y({{x}^{2}}-{{y}^{2}})$
By using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ solve the expression further.
$\Rightarrow {{x}^{2}}(x+y)-y(x-y)(x+y)$
Now, take (x + y) common:
$\Rightarrow (x+y)\left[ {{x}^{2}}-y\left( x-y \right) \right]$
Last step is to open the inner bracket:
$\Rightarrow (x+y)({{x}^{2}}-xy+{{y}^{2}})$
See, we got the exact result by different methods. We got all the factors of ${{x}^{3}}+{{y}^{3}}$. Children should know all the identities and formulae before solving this question.