Question

refer to the attachment for the Question ......



can this question be solved by some other easy method other than using sophie german identity??
and.. if yes ..plz show by some easy method too!!
​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Algebraic Identities have been used. Firstly we shall find a common term and then simplify it. Using that simplification, we will simplify each terms of numerator and denominator. We shall do it without using Sophie Germain Identity.

Let's do it !!

_____________________________________________

★ To Find :-

$\\\;\bf{\green{\dfrac{(10^{4}\:+\:324)(22^{4}\:+\:324)(34^{4}\:+\:324)(46^{4}\:+\:324)(58^{4}\:+\:324)}{(4^{4}\:+\:324)(16^{4}\:+\;324)(28^{4}\:+\:324)(40^{4}\:+\:324)(52^{4}\:+\:324)}}}$

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★ Solution :-

We see that here 324 is a common term. This can be written as,

✒ 324 = 4 × 3⁴ = 4(3⁴)

And also we see that (a⁴ + 4b⁴) is a common pattern in all terms where b⁴ is equal to 3⁴ .

This identity, can we be written as

✒ a⁴ + 4b⁴ = (a⁴ + 4b⁴ + 4a²b²) - 4a²b²

= (a² + b²)² - (2ab)²

✒ a⁴ + 4b⁴ = (a² + 2b² + 2ab)(a² + 2b² - 2ab)

Then simplifying numerator and denominator :-

_____________________________________________

~ For Elements of Numerator ::

$\\\;\tt{\gray{\rightarrow\;\;N_{1}\;=\;[10^{4}\;+\;324]}}$

→ N₁ = [10⁴ + 4(3⁴)]

→ N₁ = [10² + 2(3)² + 2(10)(3)] × [10² + 2(3)² - 2(10)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{1}\;=\;178\;\times\;58}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{2}\;=\;[22^{4}\;+\;324]}}$

→ N₂ = [22⁴ + 4(3⁴)]

→ N₂ = [22² + 2(3)² + 2(22)(3)] × [22² + 2(3)² - 2(22)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{2}\;=\;634\;\times\;370}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{3}\;=\;[34^{4}\;+\;324]}}$

→ N₃ = [34⁴ + 4(3⁴)]

→ N₃ = [34² + 2(3)² + 2(34)(3)] × [34² + 2(3)² - 2(34)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{3}\;=\;1378\;\times\;970}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{4}\;=\;[46^{4}\;+\;324]}}$

→ N₄ = [46⁴ + 4(3⁴)]

→ N₄ = [46² + 2(3)² + 2(46)(3)] × [46² + 2(3)² - 2(46)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{4}\;=\;2410\;\times\;1858}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{5}\;=\;[58^{4}\;+\;324]}}$

→ N₅ = [58⁴ + 4(3⁴)]

→ N₅ = [58² + 2(3)² + 2(58)(3)] × [58² + 2(3)² - 2(58)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{5}\;=\;3730\;\times\;3034}}$

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~ For the Elements of the Denominator ::

$\\\;\tt{\gray{\rightarrow\;\;D_{1}\;=\;[4^{4}\;+\;324]}}$

→ D₁ = [4⁴ + 4(3⁴)]

→ D₁ = [4² + 2(3)² + 2(4)(3)] × [4² + 2(3)² - 2(4)(3)]

$\\\;\tt{\orange{\leadsto\;\;D_{1}\;=\;58\;\times\;10}}$

$\\\;\tt{\gray{\rightarrow\;\;D_{2}\;=\;[16^{4}\;+\;324]}}$

→ D₂ = [16⁴ + 4(3⁴)]

→ D₂ = [16² + 2(3)² + 2(16)(3)] × [16² + 2(3)² - 2(16)(3)]

$\\\;\tt{\orange{\leadsto\;\;D_{2}\;=\;370\;\times\;178}}$

$\\\;\tt{\gray{\rightarrow\;\;D_{3}\;=\;[28^{4}\;+\;324]}}$

→ D₃ = [28⁴ + 4(3⁴)]

→ D₃ = [28² + 2(3)² + 2(28)(3)] × [28² + 2(3)² - 2(28)(3)]

$\\\;\tt{\orange{\leadsto\;\;D_{3}\;=\;970\;\times\;634}}$

$\\\;\tt{\gray{\rightarrow\;\;D_{4}\;=\;[40^{4}\;+\;324]}}$

→ D₄ = [40⁴ + 4(3⁴)]

→ D₄ = [40² + 2(3)² + 2(40)(3)] × [40² + 2(3)² - 2(40)(3)]

$\\\;\tt{\orange{\leadsto\;\;D_{4}\;=\;1858\;\times\;1378}}$

$\\\;\tt{\gray{\rightarrow\;\;D_{5}\;=\;[52^{4}\;+\;324]}}$

→ D₅ = [52⁴ + 4(3⁴)]

→ D₅ = [52² + 2(3)² + 2(52)(3)] × [52² + 2(3)² - 2(52)(3)]

$\\\;\tt{\orange{\leadsto\;\;D_{5}\;=\;3034\;\times\;2410}}$

_____________________________________________

~ For the Final Equation ::

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(10^{4}\:+\:324)(22^{4}\:+\:324)(34^{4}\:+\:324)(46^{4}\:+\:324)(58^{4}\:+\:324)}{(4^{4}\:+\:324)(16^{4}\:+\;324)(28^{4}\:+\:324)(40^{4}\:+\:324)(52^{4}\:+\:324)}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(N_{1})(N_{2})(N_{3})(N_{4})(N_{5})}{(D_{1})(D_{2})(D_{3})(D_{4})(D_{5})}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(178\;\times\;58)(634\;\times\;370)(1378\;\times\;970)(2410\;\times\;1858)(3730\;\times\;3034)}{(58\;\times\;10)(370\;\times\;178)(970\;\times\;634)(1858\;\times\;1378)(3034\;\times\;2410)}}$

Cancelling the like terms, we get,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{3730}{10}}$

$\\\;\bf{:\Longrightarrow\;\;373}$

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;answer\;\;=\;\bf{\purple{373}}}}}$

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★ More to know :-

Sophie Germain Identity ::

$\\\;\sf{\mapsto\;\;a^{4}\;+\;4b^{4}\;=\;(a^{2}\:+\:2b^{2}\:-\:2ab)(a^{2}\:+\:2b^{2}\:+\:2ab)}$

Answer 2

$\large\bold\green{\checkmark{Verified~Answer}}$

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Algebraic Identities have been used. Firstly we shall find a common term and then simplify it. Using that simplification, we will simplify each terms of numerator and denominator. We shall do it without using Sophie Germain Identity.

Let's do it !!

_____________________________________________

★ To Find :-

$\\\;\bf{\color{red}{\dfrac{(10^{4}\:+\:324)(22^{4}\:+\:324)(34^{4}\:+\:324)(46^{4}\:+\:324)(58^{4}\:+\:324)}{(4^{4}\:+\:324)(16^{4}\:+\;324)(28^{4}\:+\:324)(40^{4}\:+\:324)(52^{4}\:+\:324)}}}$

_____________________________________________

★ Solution :-

We see that here 324 is a common term. This can be written as,

✒ 324 = 4 × 3⁴ = 4(3⁴)

And also we see that (a⁴ + 4b⁴) is a common pattern in all terms where b⁴ is equal to 3⁴ .

This identity, can we be written as

✒ a⁴ + 4b⁴ = (a⁴ + 4b⁴ + 4a²b²) - 4a²b²

= (a² + b²)² - (2ab)²

✒ a⁴ + 4b⁴ = (a² + 2b² + 2ab)(a² + 2b² - 2ab)

Then simplifying numerator and denominator :-

_____________________________________________

~ For Elements of Numerator ::

$\\\;\tt{\blue{\rightarrow\;\;N_{1}\;=\;[10^{4}\;+\;324]}}$

→ N₁ = [10⁴ + 4(3⁴)]

→ N₁ = [10² + 2(3)² + 2(10)(3)] × [10² + 2(3)² - 2(10)(3)]

$\\\;\tt{\pink{\leadsto\;\;N_{1}\;=\;178\;\times\;58}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{2}\;=\;[22^{4}\;+\;324]}}$

→ N₂ = [22⁴ + 4(3⁴)]

→ N₂ = [22² + 2(3)² + 2(22)(3)] × [22² + 2(3)² - 2(22)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{2}\;=\;634\;\times\;370}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{3}\;=\;[34^{4}\;+\;324]}}$

→ N₃ = [34⁴ + 4(3⁴)]

→ N₃ = [34² + 2(3)² + 2(34)(3)] × [34² + 2(3)² - 2(34)(3)]

$\\\;\tt{\orange{\leadsto\;\;N_{3}\;=\;1378\;\times\;970}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{4}\;=\;[46^{4}\;+\;324]}}$

→ N₄ = [46⁴ + 4(3⁴)]

→ N₄ = [46² + 2(3)² + 2(46)(3)] × [46² + 2(3)² - 2(46)(3)]

$\\\;\tt{\red{\leadsto\;\;N_{4}\;=\;2410\;\times\;1858}}$

$\\\;\tt{\gray{\rightarrow\;\;N_{5}\;=\;[58^{4}\;+\;324]}}$

→ N₅ = [58⁴ + 4(3⁴)]

→ N₅ = [58² + 2(3)² + 2(58)(3)] × [58² + 2(3)² - 2(58)(3)]

$\\\;\tt{\green{\leadsto\;\;N_{5}\;=\;3730\;\times\;3034}}$

_____________________________________________

~ For the Elements of the Denominator ::

$\\\;\tt{\pink{\rightarrow\;\;D_{1}\;=\;[4^{4}\;+\;324]}}$

→ D₁ = [4⁴ + 4(3⁴)]

→ D₁ = [4² + 2(3)² + 2(4)(3)] × [4² + 2(3)² - 2(4)(3)]

$\\\;\tt{\purple{\leadsto\;\;D_{1}\;=\;58\;\times\;10}}$

$\\\;\tt{\pink{\rightarrow\;\;D_{2}\;=\;[16^{4}\;+\;324]}}$

→ D₂ = [16⁴ + 4(3⁴)]

→ D₂ = [16² + 2(3)² + 2(16)(3)] × [16² + 2(3)² - 2(16)(3)]

$\\\;\tt{\pink{\leadsto\;\;D_{2}\;=\;370\;\times\;178}}$

$\\\;\tt{\purple{\rightarrow\;\;D_{3}\;=\;[28^{4}\;+\;324]}}$

→ D₃ = [28⁴ + 4(3⁴)]

→ D₃ = [28² + 2(3)² + 2(28)(3)] × [28² + 2(3)² - 2(28)(3)]

$\\\;\tt{\purple{\leadsto\;\;D_{3}\;=\;970\;\times\;634}}$

$\\\;\tt{\pink{\rightarrow\;\;D_{4}\;=\;[40^{4}\;+\;324]}}$

→ D₄ = [40⁴ + 4(3⁴)]

→ D₄ = [40² + 2(3)² + 2(40)(3)] × [40² + 2(3)² - 2(40)(3)]

$\\\;\tt{\purple{\leadsto\;\;D_{4}\;=\;1858\;\times\;1378}}$

$\\\;\tt{\orange{\rightarrow\;\;D_{5}\;=\;[52^{4}\;+\;324]}}$

→ D₅ = [52⁴ + 4(3⁴)]

→ D₅ = [52² + 2(3)² + 2(52)(3)] × [52² + 2(3)² - 2(52)(3)]

$\\\;\tt{\pink{\leadsto\;\;D_{5}\;=\;3034\;\times\;2410}}$

_____________________________________________

~ For the Final Equation ::

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(10^{4}\:+\:324)(22^{4}\:+\:324)(34^{4}\:+\:324)(46^{4}\:+\:324)(58^{4}\:+\:324)}{(4^{4}\:+\:324)(16^{4}\:+\;324)(28^{4}\:+\:324)(40^{4}\:+\:324)(52^{4}\:+\:324)}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(N_{1})(N_{2})(N_{3})(N_{4})(N_{5})}{(D_{1})(D_{2})(D_{3})(D_{4})(D_{5})}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{(178\;\times\;58)(634\;\times\;370)(1378\;\times\;970)(2410\;\times\;1858)(3730\;\times\;3034)}{(58\;\times\;10)(370\;\times\;178)(970\;\times\;634)(1858\;\times\;1378)(3034\;\times\;2410)}}$

Cancelling the like terms, we get,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{3730}{10}}$

$\\\;\bf{:\Longrightarrow\;\;373}$

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;answer\;\;=\;\bf{\purple{373}}}}}$

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