Answer

Verified

349.5k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Three liquids are given to you One is hydrochloric class 11 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE