Question

if x/y+y/x=-1 the vLue of x^3-y^3 is

Answer 1

Answer:

Answer x³ - y³ = 0

Step-by-step explanation:

To find:

Value of x³ - y³

$\frac{x}{y} + \frac{y}{x} = - 1 \\ \\ \frac{ {x}^{2} + {y}^{2} }{xy} = - 1 \\ \\ {x}^{2} + {y}^{2} + xy = 0$

Now use by using identity

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

Now put value if x² + y²

$\\ \\ (x - y) \times ( - xy + xy) \\ \\ = (x - y)(0) \\ \\ = 0$

Answer x³ - y³ = 0

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$\boxed{\bold{\mathsf{INDENITITIES \: USED}}}$

• A³ - B³ = (A-B)×(A² + AB + B²)

Answer 2

Answer: 0

Step-by-step explanation:

Given,

$\frac{x}{y}+\frac{y}{x}=-1\\\;\\\text{Taking LCM}\\\;\\\frac{x^2+y^2}{xy}=-1\\\;\\x^2+y^2=-xy\\\;\\x^2+y^2+xy=0\;\;\;\;............i)$

We know that,

$x^3-y^3=(x-y)(x^2+y^2+xy)\\\;\\x^3-y^3=(x-y)\times0\;\;\;\;\;\;\;[from\;\;eq.\;i)]\\\;\\x^3-y^3=0$

Note:-

$1.)\;\;a^3-b^3=(a-b)(a^2+b^2+ab)\\\;\\2.)\;\;a^3+b^3=(a+b)(a^2+b^2-ab)\\\;\\3.)\;\;(a+b)^3=a^3+b^3+3ab(a+b)\\\;\\4.)\;\;(a-b)^3=a^3-b^3-3ab(a-b)$