Question

यदि x/y + y/x = -1 ( x , y not equals to 0)
तो x³ - y³ ka मान hai
a) -1
b) 1
c) 0
d) ½

with solution
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Answer 1

Required Answer:-

Given:

$\rm \mapsto \dfrac{x}{y} + \dfrac{y}{x} = - 1$

To find:

• The value of x³ - y³

Solution:

Given that,

$\rm\dfrac{x}{y} + \dfrac{y}{x} = - 1$

$\rm \implies \dfrac{ {x}^{2} + {y}^{2} }{xy} = - 1$

$\rm \implies {x}^{2} + {y}^{2} = - xy$

$\rm \implies {x}^{2} + xy + {y}^{2} =0$

Now,

$\rm {x}^{3} - {y}^{3}$

$\rm = (x - y)( {x}^{2} + xy + {y}^{2} )$

As x² + xy + y² = 0, So,

$\rm (x - y)( {x}^{2} + xy + {y}^{2} )$

$\rm (x - y)\times 0$

As any number multiplied by 0 is 0, So,

Result = 0.

So,

$\rm \implies{x}^{3} - {y}^{3} = 0$

Option C is the answer for your question.

Answer:

$\rm \implies{x}^{3} - {y}^{3} = 0$

Identity Used:

➡ x³ - y³ = (x - y)(x² + xy + y²)

Other Identities:

➡ (x + y)² = x² + 2xy + y²

➡ (x - y)² = x² - 2xy + y²

➡ x² - y² = (x + y)(x - y)

➡ (x + y)² + (x - y)² = 2(x² + y²)

➡ (x + y)² - (x - y)² = 4xy

➡ x³ + y³ = (x + y)(x² - xy + y²)

Answer 2

Answer:

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