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# Factorise the polynomial ${{x}^{6}}-7{{x}^{3}}-8$

Last updated date: 21st Sep 2024
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Hint: In the given polynomial ${{x}^{6}}-7{{x}^{3}}-8$, the highest power of $x$ is equal to $6$. So this means that the degree of the polynomial is $6$. So we first have to simplify it by substituting ${{x}^{3}}=y$ so that we will obtain the quadratic polynomial ${{y}^{2}}-7y-8$ which can be easily factored using the middle term splitting method. Then, we have to back substitute $y={{x}^{3}}$ and use the algebraic identities ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ and ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ to further factorize the obtained polynomial.

Complete step-by-step solution:
Let us write the polynomial given in the question as
$p\left( x \right)={{x}^{6}}-7{{x}^{3}}-8$
Since the highest power of $x$ in the given polynomial is equal to $6$, so the degree of the given polynomial is equal to $6$.
Now, the above polynomial can also be written as
$\Rightarrow p\left( x \right)={{\left( {{x}^{3}} \right)}^{2}}-7{{x}^{3}}-8$
For simplifying the given polynomial, we substitute ${{x}^{3}}=y$ in the given polynomial to get
$\Rightarrow p\left( y \right)={{y}^{2}}-7y-8$
Now, we have a quadratic polynomial, which can be factored by using the middle term splitting method. For this, we need to split the middle term $-7y$ as the sum of two terms whose product is equal to the product of the first and the third terms, that is $-8{{y}^{2}}$. So we split the middle term as $-7y=y-8y$ in the above equation to get
$\Rightarrow p\left( y \right)={{y}^{2}}+y-8y-8$
Now, taking $y$ common from the first two terms and $-8$ common from the last two terms we get
$\Rightarrow p\left( y \right)=y\left( y+1 \right)-8\left( y+1 \right)$
We can take $\left( y+1 \right)$ common to get
$\Rightarrow p\left( y \right)=\left( y+1 \right)\left( y-8 \right)$
Now, according to our substitution, $y={{x}^{3}}$. Putting this above, we get
$\Rightarrow p\left( x \right)=\left( {{x}^{3}}+1 \right)\left( {{x}^{3}}-8 \right)$
Now, we know that $8={{2}^{3}}$, and $1={{1}^{3}}$. So we can write the above polynomial as
$\Rightarrow p\left( x \right)=\left( {{x}^{3}}+{{1}^{3}} \right)\left( {{x}^{3}}-{{2}^{3}} \right)$
Now, we know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ and ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. So the above polynomial can be written as
\begin{align} & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x+{{1}^{2}} \right)\left( x-2 \right)\left( {{x}^{2}}+2x+{{2}^{2}} \right) \\ & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right) \\ & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( x-2 \right)\left( {{x}^{2}}-x+1 \right)\left( {{x}^{2}}+2x+4 \right) \\ \end{align}
Hence, the given polynomial is factored completely.

Note: In the final factored polynomial obtained in the above polynomial, we can see that we have quadratic factors. Do not think of factorising them into linear factors since they cannot be factorised further. We can check whether a given quadratic factor can be factored further or not by checking the value of its discriminant, given by $D={{b}^{2}}-4ac$. If the discriminant is negative, this means that the quadratic factor cannot be factorized.