Question

If x = a + 1/a and y = a - 1/a then find the value of x⁴ + y⁴ - 2x²y²​

Answer 1

Answer:

x⁴ + y⁴ - 2x²y² = (16 / a⁴)

Explanation:

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Answer 2

Given :-

• $\sf x = a + \dfrac{1}{a} \: \: , \: \: y = a - \dfrac{1}{a}$

To Find:-

• The value of $\sf x⁴ + y⁴ - 2x²y²$

Formula used:-

• $\sf{(a+b)²=\green{{\underline{\underline{\pmb{a^{2}+2ab+b^{2}}}}}}}$

• $\sf{(a-b)²=\green{{\underline{\underline{\pmb{a^{2}-2ab+b^{2}}}}}}}$

Solution:-

We are given :-

• $\sf \green{x = a + \dfrac{1}{a}}$

• $\sf\green{ y = a - \dfrac{1}{a}}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━⠀

$\sf\pink{ \: \: \: \: \: \: \: \: {x}^{4} + {y}^{4} - 2 {x}^{2} {y}^{2} = ( {x}^{2} - {y}^{2})^{2}}$

$\sf\: \: \: \: \: \: \: \:: \implies ( {x}^{2} - {y}^{2})^{2}$

$\sf \: \: \: \: \: \: \: \::\implies \Bigg[ { \bigg(a + \dfrac{1}{a}} \bigg)^{2} -{\bigg(a - \dfrac{1}{a}} \bigg)^{2} \Bigg] ^{2}$

$\sf \: \: \: \: \: \: \: \::\implies \Bigg[ { \bigg( {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 {(a}^{2} ) \times \dfrac{1}{ {a}^{2} } } \bigg) -{\bigg( {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 {(a}^{2} ) \times \dfrac{1}{ {a}^{2} } } \bigg) \bigg] ^{2}$

$\sf \: \: \: \: \: \: \: \::\implies \Bigg[ { \bigg( {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 } \bigg) -{\bigg( {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 } \bigg) \bigg] ^{2}$

$\sf \: \: \: \: \: \: \: \::\implies \Bigg[ { {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 } -{ {a}^{2} - \dfrac{1}{ {a}^{2} } + 2 } \bigg] ^{2}$

$\sf \: \: \: \: \: \: \: \::\implies \Bigg[ { {a}^{2} - {a}^{2} + \dfrac{1}{ {a}^{2} } - \dfrac{1}{ {a}^{2} } + 2+ 2 } \bigg] ^{2}$

$\sf \: \: \: \: \: \: \: \::\implies {(2 + 2)}^{2}$

$\sf \: \: \: \: \: \: \: \pink{\::\implies {4}^{2} = 16 }$

$\therefore \: \underline{\textsf{\textbf{ Value \: of \:x⁴ + y⁴ - 2x²y² \: is \underline{\: 16} }}}$