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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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## 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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