(x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)(x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right) is equal to ?" /> (x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)(x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right) is equal to ?" /> (x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)(x+1x)(x−1x)(x2+1x2−1)(x2+1x2+1)\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right) is equal to ?" />

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$(x + \frac{1}{x}) (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) (x^{2} + \frac{1}{x^{2}} + 1)$ $\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$ is equal to ?

1 $x^{6} - \frac{1}{x^{6}}$ ${x^6} - \frac{1}{{{x^6}}}$
2 $x^{8} - \frac{1}{x^{8}}$ ${x^8} - \frac{1}{{{x^8}}}$
3 $x^{6} + \frac{1}{x^{6}}$ ${x^6} + \frac{1}{{{x^6}}}$
4 $x^{8} + \frac{1}{x^{8}}$ ${x^8} + \frac{1}{{{x^8}}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression , \\ [(x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - x . \frac{1}{x})] [(x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} x . \frac{1}{x})] \\ = (x^{3} + \frac{1}{x^{3}}) (x^{3} + \frac{1}{x^{3}}) \\ = x^{6} + \frac{1}{x^{6}} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \,\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}x.\frac{1}{x}} \right)} \right] \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = {x^6} + \frac{1}{{{x^6}}} \cr}$

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is equal to ?", "text": "

(x + \frac{1}{x}) (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) (x^{2} + \frac{1}{x^{2}} + 1)

$\left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$ is equal to ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression , \\ [(x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - x . \frac{1}{x})] [(x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} x . \frac{1}{x})] \\ = (x^{3} + \frac{1}{x^{3}}) (x^{3} + \frac{1}{x^{3}}) \\ = x^{6} + \frac{1}{x^{6}} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \,\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}x.\frac{1}{x}} \right)} \right] \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = {x^6} + \frac{1}{{{x^6}}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given expression , \\ [(x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - x . \frac{1}{x})] [(x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} x . \frac{1}{x})] \\ = (x^{3} + \frac{1}{x^{3}}) (x^{3} + \frac{1}{x^{3}}) \\ = x^{6} + \frac{1}{x^{6}} \end{aligned}

$\eqalign{ & {\text{Given expression ,}} \cr & \,\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}x.\frac{1}{x}} \right)} \right] \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = {x^6} + \frac{1}{{{x^6}}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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