Question

If x−y = 15 and xy = 20, then find the value of 1/x^3 −1/y^3 .​

Answer 1

Given : x−y = 15 and xy = 20

To Find :  1/x³ - 1/y³

Solution:

x−y = 15

Squaring both sides

( x - y)² = 15²

=> x² + y² - 2xy  = 225

=> x² + y² - 2(20) = 225

=> x² + y² = 265

1/x³ - 1/y³

= (y³ - x³)/x³y³

= ( y - x) ( y² + x²  + xy)/(xy)³

y - x= -(x-y) = -15

y² + x² = 265

xy = 20

= (-15) (265 + 20)/(20)³

= - 15(285) /20³

=  - 4,275/8000

= -171/320

= -0.534375

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Answer 2

SOLUTION

GIVEN

x − y = 15 and xy = 20

TO DETERMINE

$\displaystyle \sf{ \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } }$

EVALUATION

Here it is given that

x − y = 15 and xy = 20

So

$\displaystyle \sf{ \frac{x - y}{xy} = \frac{15}{20} }$

$\displaystyle \sf{ \implies \frac{1}{y} - \frac{1}{x} = \frac{3}{4} }$

$\displaystyle \sf{ \implies \frac{1}{x} - \frac{1}{y} = - \frac{3}{4} }$

Now cubing both sides we get

$\displaystyle \sf{ {\bigg( \frac{1}{x} - \frac{1}{y} \bigg)}^{3} = - \frac{27}{64} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } - 3 \times \frac{1}{x} \times \frac{1}{y} {\bigg( \frac{1}{x} - \frac{1}{y} \bigg)}^{} = - \frac{27}{64} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } - \frac{3}{xy} {\bigg( \frac{1}{x} - \frac{1}{y} \bigg)}^{} = - \frac{27}{64} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } + \frac{3}{20} \times \frac{3}{4} = - \frac{27}{64} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } = - \frac{27}{64} - \frac{9}{80} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } = - \frac{270 + 72}{640} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } = - \frac{342}{640} }$

$\displaystyle \sf{ \implies \: \frac{1}{ {x}^{3} } - \frac{1}{ {y}^{3} } = - \frac{171}{320} }$

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