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

Last updated date: 22nd Jul 2024
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Hint: While factorising the given quadratic equation of the form $a{{x}^{2}}+bx+c$, We have to choose two numbers such that if we multiply those two numbers, we have to get $ac$and if we add, we have to get $b$. In this question we will use some basic algebraic formulas ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.

Complete step-by-step solution:
From the question it is clear that we have to factorise ${{x}^{6}}-7{{x}^{3}}-8$.
Now let us consider ${{x}^{6}}-7{{x}^{3}}-8$ as $y$
$\Rightarrow y={{x}^{6}}-7{{x}^{3}}-8$……………..(1)
In the equation ${{x}^{6}}-7{{x}^{3}}-8$ we can see the highest degree as $6$. We can write $6$ as $3\times 2$.
$\Rightarrow y={{x}^{3\times 2}}-7{{x}^{3}}-8$
From the basic rules of mathematics, we can write ${{x}^{a\times b}}={{\left( {{x}^{3}} \right)}^{2}}$. So, we can write ${{x}^{3\times 2}}={{\left( {{x}^{3}} \right)}^{2}}$
$\Rightarrow y={{\left( {{x}^{3}} \right)}^{2}}-7{{x}^{3}}-8$…………………(2)
Now let us assume ${{x}^{3}}$ is equal to $a$
So, equation (2) becomes as
$\Rightarrow y={{\left( a \right)}^{2}}-7\times a-8$…………………….(3)
It is quadratic equation in the form of $a{{x}^{2}}+bx+c$, while factorising the quadratic equation, $a{{x}^{2}}+bx+c$, we have to choose two numbers such that if we multiply those two numbers, we have to get $ac$and if we add, we have to get $b$.
Now let us try to rewrite the equation (3) by writing $-7a$ as $-8a+a$.
$\Rightarrow y={{\left( a \right)}^{2}}-8a+a-8$
Consider the terms ${{\left( a \right)}^{2}}-8a$from the equation. Take out $a$ common from ${{\left( a \right)}^{2}}-8a$
$\Rightarrow y=a\left( a-8 \right)+\left( a-8 \right)$
Now take out common $\left( a-8 \right)$ from $a\left( a-8 \right)+\left( a-8 \right)$.
We get,
$\Rightarrow y=\left( a-8 \right)\left( a+1 \right)$……………….(4)
Since we have taken ${{x}^{3}}=a$, put $a={{x}^{3}}$in the equation (4)
$\Rightarrow y=\left( {{x}^{3}}-8 \right)\left( {{x}^{3}}+1 \right)$
From the standard cubic values, we can write $8={{2}^{3}}$ and $1={{1}^{3}}$
$\Rightarrow y=\left( {{x}^{3}}-{{2}^{3}} \right)\left( {{x}^{3}}+{{1}^{3}} \right)$……………….(5)
From the algebraic formulas we know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. Using these two identities, we can write
$\Rightarrow \left( {{x}^{3}}-{{2}^{3}} \right)=(x-2)({{x}^{2}}+2x+4)$
$\Rightarrow \left( {{x}^{3}}+{{1}^{3}} \right)=(x+1)({{x}^{2}}-x+1)$
Put these two equations in equation (6)
$\Rightarrow y=\left( {{x}^{3}}-{{2}^{3}} \right)\left( {{x}^{3}}+{{1}^{3}} \right)$
$\Rightarrow y=(x-2)({{x}^{2}}+2x+4)\times (x+1)({{x}^{2}}-x+1)$
Hence $(x-2)({{x}^{2}}+2x+4)\times (x+1)({{x}^{2}}-x+1)$ is the final answer.

Note: Students should avoid mistakes while solving this problem. Students should use correct algebraic formulas. In case if there is use of wrong formulas then the final answer may get interrupted. So, students should avoid these mistakes while solving this problem such that the final answer can be obtained in a correct manner.