How to solve 8(x+1/x)^2+4(x^2+1/x^2)^2-4(x^2+1/x^2)(x+1/x)^2=(x+4)^2 ?

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Welcome to my article 8(x+1/x)^2+4(x^2+1/x^2)^2-4(x^2+1/x^2)(x+1/x)^2=(x+4)^2 .
This article is taken from the simplification lesson, in this article we have been told how to solve the problem easily by doing addition, subtraction, multiplication, division and fractionation.or complete information on how to solve this 1/x+1+2/x+2=5/x+4, read and understand it carefully.

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First of all we should write the article on the page of the notebook.

8(x+1/x)^2+4(x^2+1/x^2)^2-4(x^2+1/x^2)(x+1/x)^2=(x+4)^2 ,

To solve this question write in this way,

\displaystyle 8{{\left( {x+\frac{1}{x}} \right)}^{2}}+4{{\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}} \right)}^{2}}-4\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}} \right).{{\left( {x+\frac{1}{x}} \right)}^{2}}={{(x+4)}^{2}}

\displaystyle 8\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}+2} \right)+4\left( {{{x}^{4}}+\frac{1}{{{{x}^{4}}}}+2} \right)-4\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}} \right).\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}+2} \right)={{(x+4)}^{2}}

\displaystyle 8\left( {{{x}^{2}}+\frac{1}{{{{x}^{2}}}}+2} \right)+4\left( {{{x}^{4}}+\frac{1}{{{{x}^{4}}}}+2} \right)-4\left[ {{{x}^{4}}+1+2{{x}^{2}}+1+\frac{1}{{{{x}^{4}}}}+\frac{2}{{{{x}^{2}}}}} \right]={{(x+4)}^{2}}

\displaystyle 8{{x}^{2}}+\frac{8}{{{{x}^{2}}}}+16+4{{x}^{4}}+\frac{4}{{{{x}^{4}}}}+8-4{{x}^{4}}-4-8{{x}^{2}}-4-\frac{4}{{{{x}^{4}}}}-\frac{8}{{{{x}^{2}}}}={{(x+4)}^{2}}

\displaystyle 8{{x}^{2}}-8{{x}^{2}}+\frac{8}{{{{x}^{2}}}}-\frac{8}{{{{x}^{2}}}}+4{{x}^{4}}-4{{x}^{4}}+\frac{4}{{{{x}^{4}}}}-\frac{4}{{{{x}^{4}}}}+8-4-4+16={{(x+4)}^{2}}

\displaystyle 8-8+16={{(x+4)}^{2}}

\displaystyle 16={{(x+4)}^{2}}

\displaystyle 16={{x}^{2}}+{{4}^{2}}+2.x.4

\displaystyle 16={{x}^{2}}+16+8x

\displaystyle 16-{{x}^{2}}-16-8x=0

\displaystyle -{{x}^{2}}-8x=0

\displaystyle -({{x}^{2}}+8x)=-(0)

\displaystyle {{x}^{2}}+8x=0

\displaystyle x(x+8)=0

\displaystyle x+8=0

\displaystyle x=-8Answer

The formula used in this question,

\displaystyle {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab

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