Question

answer this question.​

Answer 1

hope you understand the answer

Answer 2

Step-by-step explanation:

Question:-

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

Given:-

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

To Find:-

• The value of x⁴ + y⁴ - 2x²y²

Formula used:-

• (x + y)² = x² + y² + 2ab

• (x - y)² = x² + y² - 2ab

Solution:-

x⁴ + y⁴ - 2x²y² is in the form x² + y² - 2xy

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

• $\sf x = a + \dfrac{1}{a}$

• $\sf y = a - \dfrac{1}{a}$

$\:$

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

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

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

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

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

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

$\sf = {(2 + 2)}^{2}$

$\sf = {4}^{2} = 16$

$\dashrightarrow \large\underline{\boxed{\sf \therefore {x}^{4} + {y}^{4} - 2 {x}^{2} {y}^{2} = 16}}$

$\star \: \underline{\textsf{\textbf{ The \: Answer \: is\: Option \: (c)}}}$