Question 13

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Maths Question: 

$\begin{align}  & \text{Let }x={{(a+b)}^{\tfrac{1}{4}}}+{{(a-b)}^{\tfrac{1}{4}}},\text{ }y={{(a+b)}^{\tfrac{1}{4}}}-{{(a-b)}^{\tfrac{1}{4}}}\text{ and } \\ & {{a}^{2}}-{{b}^{2}}={{p}^{4}};\text{ prove }{{x}^{2}}-4p-{{y}^{2}}=0 \\\end{align}$

Maths Solution: 

$\begin{align}  & x={{(a+b)}^{\tfrac{1}{4}}}+{{(a-b)}^{\tfrac{1}{4}}} \\ & {{x}^{2}}={{\left[ {{(a+b)}^{\tfrac{1}{4}}}+{{(a-b)}^{\tfrac{1}{4}}} \right]}^{2}} \\ & {{x}^{2}}={{(a+b)}^{\tfrac{1}{2}}}+{{(a-b)}^{\tfrac{1}{2}}}+2{{(a+b)}^{\tfrac{1}{4}}}{{(a-b)}^{\tfrac{1}{4}}}---(i) \\ & y={{(a+b)}^{\tfrac{1}{4}}}-{{(a-b)}^{\tfrac{1}{4}}} \\ & {{y}^{2}}={{\left[ {{(a+b)}^{\tfrac{1}{4}}}-{{(a-b)}^{\tfrac{1}{4}}} \right]}^{2}} \\ & {{y}^{2}}={{(a+b)}^{\tfrac{1}{2}}}+{{(a-b)}^{\tfrac{1}{2}}}-2{{(a+b)}^{\tfrac{1}{4}}}{{(a-b)}^{\tfrac{1}{4}}}---(ii) \\ & \text{Subtract }(ii)\text{ from }(i) \\ & {{x}^{2}}-{{y}^{2}}=4{{(a+b)}^{\tfrac{1}{4}}}{{(a-b)}^{\tfrac{1}{4}}}----(iii) \\ & \text{Since }{{p}^{4}}={{a}^{2}}-{{b}^{2}} \\ & {{p}^{4}}=(a-b)(a+b) \\ & \text{Find the fourth root of both side} \\ & p={{(a-b)}^{\tfrac{1}{4}}}{{(a+b)}^{\tfrac{1}{4}}} \\ & \text{Substitute }{{(a-b)}^{\tfrac{1}{4}}}{{(a+b)}^{\tfrac{1}{4}}}\text{ for }p\text{ in }(iii) \\ & {{x}^{2}}-{{y}^{2}}=4p \\ & {{x}^{2}}-{{y}^{2}}=4p \\ & {{x}^{2}}-4p-{{y}^{2}}=0\text{    (}proved) \\\end{align}$ 

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Comments

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