Question

Given : ${\sf{x^4 + x^2 y^2 + y^4 = 189}}$

${\sf{x^2 + xy + y^2 = 21}}$

To Find : ${\sf{\ \ {\dfrac{1}{x^2}} + {\dfrac{1}{y^2}} }}$
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Solve this..
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Answer 1

||✪✪ QUESTION ✪✪||

if x⁴ + x²y² + y⁴ = 189 and x^2 + xy + y^2 = 21, Find (1/x² + 1/y²) ?

✫✫ Solution (1) ✫✫ :-

Let,,

➪ x⁴ + x²y² + y⁴ = 189 ----- Equation (❶)

➪x² + xy + y² = 21 ------ Equation (❷)

Adding and subtracting x²y² in Equation ❶ LHS we get,

➺ x⁴ + x²y² + y⁴ + x²y² - x²y² = 189

➺ (x⁴ + 2x²y² + y⁴) - x²y² = 189

Now, using a² + b² + 2ab = (a+b)² , we get,

➺ (x² + y²)² - (xy)² = 189

Now, using (a² - b²) = (a+b)(a-b),

➺ (x² + y² + xy)(x² + y² - xy) = 189

Putting value of Equation ❷ here , we get,

➺ 21 * (x²+y² - xy) = 189

Dividing both sides by 21,

➺ (x² + y² - xy) = 9 ----- Equation ❸

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Now, Adding Equation ❷ & ❸ we get,

➻ (x² + y² + xy) + (x² + y² - xy) = 21 + 9

➻ 2x² + 2y² = 30

➻ 2(x² + y²) = 30

➻ (x² + y²) = 15 ------- Equation ❹

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Now, Subtracting Equation ❹ From Equation ❷ we get,

➼ (x² + y² + xy) - (x²+y²) = 21 - 15

➼ xy = 6 ------ Equation ❺

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Now, we have To Find :-

☛ (1/x²) + (1/y²)

☛ [ (y² + x²) / x²y² ]

☛ [ (x² + y²) / (xy)² ]

Putting value of Equation ❹ & ❺ Now, we get,

☛ [ 15/(6)² ]

☛ (15/36)

☛ (5/12) (Ans.)

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☙☘ Solution (2) ☘☙ :-

Let ,

1/x² + 1/y² = M

⟿ x² +y² = M(xy)² = (21- xy)

Also ,

⟼x⁴ + y⁴ + 2(xy)² = M².(xy)⁴

⟼ 189 + (xy)² = 441 + (xy)² - 42xy

⟼ xy = 6

⟼ M = (21-6)/36 = 5/12 (Ans.)

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Answer 2

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❏ Question:-

${\sf{x^4 + x^2 y^2 + y^4 = 189}}$

${\sf{x^2 + xy + y^2 = 21}}$

To Find : ${\sf{\ \ {\dfrac{1}{x^2}} + {\dfrac{1}{y^2}} }}$

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❏ Solution:-

$\star\:\:\:\:{\sf{x^2+ x y + y^2 = 21}}$........(i)

$\implies\large{{\sf{x^2 + y^2 = 21-xy}}}$

And ,

$\star\:\:\:\:{\sf{x^4 + x^2 y^2 + y^4 = 189}}$.......(ii)

$\implies{\sf{x^2 y^2 = 189-(x^4+y^4)}}$

$\implies{\sf{x^2 y^2 = 189-\{(x^2+y^2)^2-2x^2y^2\}}}$

$\implies{\sf{x^2 y^2 = 189-\{(21-xy)^2-2x^2y^2\}}}$

$\implies{\sf{x^2 y^2 = 189-\{21^2-42xy+x^2y^2-2x^2y^2\}}}$

$\implies{\sf{x^2 y^2 = 189-\{441-42xy-x^2y^2\}}}$

$\implies{\sf{x^2 y^2 = 189-441+42xy+x^2y^2}}$

$\implies{\sf{\cancel{x^2 y^2}- \cancel{x^2 y^2} = -252+42xy}}$

$\implies{\sf{{42xy=252}}}$

$\implies\large{\boxed{{\sf{xy=6}}}}$

Hence,

$\implies{\sf{\ \ {\dfrac{1}{x^2}} + {\dfrac{1}{y^2}} }}$

$\implies{\sf{\ \ { \dfrac{y^2+x^2}{x^2y^2}}}}$

$\implies{\sf{\ \ { \dfrac{21-xy}{(xy)^2}}}}$

( putting the value of xy=6)

$\implies{\sf{\ \ { \dfrac{21-6}{(6)^2}}}}$

$\implies{\sf{\ \ { \dfrac{\cancel{15}}{\cancel{36}}}}}$

$\implies\large{{\sf{\ \ { \dfrac{5}{12}}}}}$ (Ans).

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