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Question No. 7. Find out selling anddistribution Exp.Advertisement = Rs. 1000per month , Salary ofsalesmen = 90220 Wages ofshopkeeper = 8880​

Answer»

\sf{(x+y)^2={\underline{\underline{\pmb{x^{2}+2xy+y^{2}}}}}}

\sf{(x+y)(x-y)={\underline{\underline{\pmb{x^{2}-y^{2}}}}}}

\large\underline{\sf{Solution-}}

\sf :\impiles \red{\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 1}

\sf \: = \: \: \dfrac{ {x}^{8} + 1 + {x}^{4} }{ {x}^{4} }

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{8} + 2{x}^{4} + 1 - {x }^{4} \bigg)

\: \: \: \: \: \: \: \: \: \: \: \pink{ \because \sf \: On \: adding \: and \: subtracting \: {x}^{4} }

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {( {x}^{4} + 1) }^{2} - { ({x}^{2}) }^{2} \bigg)

\: \: \: \: \:

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {x}^{2} \bigg)

\: \: \: \: \: \: \: \: \:

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {2x}^{2} - {x}^{2} \bigg)

\: \: \: \: \: \: \: \: \: \: \: \pink{\because \sf \: On \: adding \: and \: subtracting \: {x}^{2} }\\\\

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg)\bigg( {( {x}^{2} + 1) }^{2} - {(x)}^{2} \bigg) \\\\

\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg(( {x}^{2} + 1 + x)( {x}^{2} + 1 - x) \bigg)\\\\

\sf \: = \: \: \bigg( \dfrac{ {x}^{4} + 1 - {x}^{2} }{ {x}^{2} } \bigg) \bigg( \dfrac{ {x}^{2} + x + 1 }{x} \bigg) \bigg( \dfrac{ {x}^{2} - x + 1}{x} \bigg)\\\\

\sf\red{ \: = \:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } - 1 \bigg) \bigg( x + \dfrac{1}{x} + 1 \bigg) \bigg(x + \dfrac{1}{x} - 1 \bigg)} \\\\

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