Question

If (x+y)^2-xy=0, then find the value of (x^3-y^3)/(x-y).
A. 2
B. 3
C. 0
D. 5​

Answer 1

Step-by-step explanation:

✏See Attachment .

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Answer 2

Answer:

$\large\bold\red{(C)0}$

Step-by-step explanation:

Given,

${(x + y)}^{2} - xy = 0 \\ \\ = > {x}^{2} + {y}^{2} + 2xy - xy = 0 \\ \\ = > {x}^{2} + xy + {y}^{2} = 0 \: \: \: \: \: \: ..............(i)$

Now,

we have to find the value of,

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

But,

we know that,

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

Therefore,

putting the values,

we get,

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

But,

from Equation (i),

we get,

${x}^{2} + {y}^{2} + xy = 0$

therefore,

putting the values,

we get,

The correct answer is,

Option $\bold{(C)0}$