Question

How do you factorize the given polynomial ${{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1$?

Answer 1

Hint: We start solving the problem by writing all the terms in the polynomial as a multiplication of two terms. We then take the common terms present in the polynomial after multiplication to problems. We then make use of the result ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to proceed through the problem. We then make use of the results ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ to proceed through the problem. We then make the necessary arrangements in the obtained result to get the required result.

Complete step by step answer:
According to the problem, we are asked to factorize the given polynomial ${{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1$.
We have given the polynomial ${{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1$.
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( {{x}^{6}}\times {{x}^{3}} \right)+\left( {{x}^{6}}\times -1 \right)+\left( -1\times {{x}^{3}} \right)+\left( -1\times -1 \right)$ ---(1).
Now, let us take out the common factor from the equation (1) to find the other factors of the given polynomial.
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1={{x}^{6}}\times \left( {{x}^{3}}-1 \right)-1\times \left( {{x}^{3}}-1 \right)$ ---(2).
Let us take the common terms present in equation (2).
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( {{x}^{6}}-1 \right)\left( {{x}^{3}}-1 \right)$.
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( {{\left( {{x}^{3}} \right)}^{2}}-{{1}^{2}} \right)\left( {{x}^{3}}-1 \right)$ ---(3).
We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Let us use this result in equation (3).
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( {{x}^{3}}+1 \right)\left( {{x}^{3}}-1 \right)\left( {{x}^{3}}-1 \right)$.
$\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( {{x}^{3}}+{{1}^{3}} \right)\left( {{x}^{3}}-{{1}^{3}} \right)\left( {{x}^{3}}-{{1}^{3}} \right)$ ---(4).
We know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. Let us use these results in equation (4).
\[\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}+{{1}^{2}}-\left( x \right)\left( 1 \right) \right)\left( x-1 \right)\left( {{x}^{2}}+{{1}^{2}}+\left( x \right)\left( 1 \right) \right)\left( x-1 \right)\left( {{x}^{2}}+{{1}^{2}}+\left( x \right)\left( 1 \right) \right)\].
\[\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\].
\[\Rightarrow {{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1=\left( x+1 \right)\left( {{x}^{2}}-x+1 \right){{\left( x-1 \right)}^{2}}{{\left( {{x}^{2}}+x+1 \right)}^{2}}\].
So, we have found the factors of the polynomial ${{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1$ as \[\left( x+1 \right)\], \[\left( {{x}^{2}}-x+1 \right)\], \[\left( x-1 \right)\], \[\left( x-1 \right)\], \[\left( {{x}^{2}}+x+1 \right)\], \[\left( {{x}^{2}}+x+1 \right)\].

$\therefore $ The factorization of the given polynomial ${{x}^{9}}-{{x}^{6}}-{{x}^{3}}+1$ is \[\left( x+1 \right)\left( {{x}^{2}}-x+1 \right){{\left( x-1 \right)}^{2}}{{\left( {{x}^{2}}+x+1 \right)}^{2}}\].

Note: Whenever we get this type of problems, we first try to find the common multiples (factors) in all the terms present in the polynomial. We should keep in mind that we need to get the same polynomial after multiplying the factors obtained from factorization. We should make calculation mistakes while solving this problem. Similarly, we can expect problems to find the factorization of the polynomial ${{x}^{8}}-{{x}^{4}}-{{x}^{2}}+1$.