# 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 .

## 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

## The formula used in this question,

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

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