Question

x/y+y/z=-1 where x,y not equal 0 then the value of (x at rate 3-y at rate3 is

Answer 1

Correct Question :

›»› x/y + y/x = -1 where x, y not equal 0 then the value of x³ - y³ is?

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Answer :

›»› The value of x³ - y³ = 0

Given :

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

To Find :

• The value of x³ - y³ = ?

Required Solution :

$\tt{:\implies \dfrac{x}{y} + \dfrac{y}{x} = - 1}$

$\tt{:\implies \dfrac{x \times x}{x \times y} + \dfrac{y \times y}{y \times x} = - 1}$

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

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

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

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

Now ,

We know that

$\tt{:\implies {x}^{3} - {y}^{3} = (x - y)( {x}^{2} \times xy + {y}^{2} )}$

$\tt{:\implies {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} \times xy)}$

$\tt{:\implies {x}^{3} - {y}^{3} = (x - y)( - \cancel{x}y + \cancel{x}y)}$

$\tt{:\implies {x}^{3} - {y}^{3} =(x - y)(0)}$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: {x}^{3} - {y}^{3} = 0 \: \: }} \: \: }}$

║Hence, the value of x³ - y³ is 0.║

Answer 2

Answer:

Given that:

$\frac{x}{y} + \frac{y}{z} = - 1$

where x and y is not equal to 0.

To find:

x^3 – y^3

We know that,

${x}^{3} - {y}^{3} \\ = (x - y)( {x}^{2} + xy + {y}^{2})$

$= > \frac{x}{y} + \frac{y}{z} = - 1 \\ \\ = > \frac{ {x}^{2} + {y}^{2} }{xy} = - 1$

Now after cross multiplying it, we get

${x}^{2} + {y}^{2} = - xy \\ \\ = > {x}^{2} + xy + {y}^{2} = 0$

Therefore,

${x}^{3} - {y}^{3} \\ \\ = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ \\ = x - y \\ \\ = 0$

Hence, x^3 – y^3 = 0.