Question

simplify. x+y÷x-y-x-y+4xy÷x^2+y^2-8x^3y÷x^4+y^4​

Answer 1

Question -

Simplify -

$\frac{x + y}{x - y} - \frac{x - y}{x + y} + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8 {x}^{3}y }{ {x}^{4} + {y}^{4} }$

Solution -

$= \frac{x + y}{x - y} - \frac{x - y}{x + y} + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8x {}^{3}y }{ {x}^{4} + {y}^{4} }$

$= \frac{(x + y)(x + y) - (x - y)(x - y)}{(x - y)(x + y)} + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8 {x}^{3} y}{ {x}^{4} + {y}^{4} }$

$= \frac{ {(x + y)}^{2} - {(x - y)}^{2} }{ {x}^{2} - {y}^{2} } + \frac{4xy}{ {x}^{2} - {y}^{2} } - \frac{8 {x}^{3}y }{ {x}^{4} + {y}^{4} }$

$= \frac{ {x}^{2} + {y}^{2} + 2xy - ( {x}^{2} + {y}^{2} - 2xy) }{ {x}^{2} - {y}^{2} } + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8 {x}^{3}y }{ {x}^{4} + {y}^{4} }$

$= \frac{ {x}^{2} + {y}^{2} + 2xy - {x}^{2} - {y}^{2} + 2xy }{ {x}^{2} - {y}^{2} } + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8 {x}^{3}y }{ {x}^{4} + {y}^{4} }$

$= \frac{4xy}{ {x}^{2} - {y}^{2} } + \frac{4xy}{ {x}^{2} + {y}^{2} } - \frac{8 {x}^{3} y}{ {x}^{4} + {y}^{4} }$

$= \frac{4xy( {x}^{2} + {y}^{2} ) + 4xy( {x}^{2} - {y}^{2}) }{( {x}^{2} - {y}^{2})( {x}^{2} + {y}^{2} ) } - \frac{8 {x}^{3} y}{ {x}^{4} + {y}^{4} }$

$= \frac{4 {x}^{3}y + 4x {y}^{3} + 4 {x}^{3}y - 4x {y}^{3} }{ {x}^{4} - {y}^{4} } - \frac{8 {x}^{3}y }{ {x}^{4} + {y}^{4} }$

$= \frac{8 {x}^{3}y }{ {x}^{4} - {y}^{4} } - \frac{8 {x}^{3}y }{ {x}^{4} - {y}^{4} }$

$= \frac{8 {x}^{3} y( {x}^{4} - {y}^{4}) - 8 {x}^{3} y( {x}^{4} - {y}^{4} )}{ {x}^{8} - {y}^{8} }$

$= \frac{8 {x}^{7} y + 8 {x}^{3} {y}^{5} - 8 {x}^{7}y + 8 {x}^{3} {y}^{5} }{ {x}^{8} - {y}^{8} }$

$= \frac{16 {x}^{3} {y}^{5} }{ {x}^{8} - {y}^{8} }$