Question
The simplified value of $$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right)$$  $$ ÷ $$ $$\left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$   is?
Answer: Option B
$$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right) \div \left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$
$$ = \left( {\frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}} \right) \div $$ Â Â $$\left( {\frac{{{x^3} - {y^3} - 3xy\left( {x - y} \right)}}{{x - y}}} \right)$$
$$\eqalign{
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div \frac{{{{\left( {x - y} \right)}^3}}}{{x - y}} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div {\left( {x - y} \right)^2} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \times \frac{1}{{{{\left( {x - y} \right)}^2}}} \cr
& = \frac{1}{{{x^2} + {y^2}}} \cr} $$
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$$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right) \div \left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$
$$ = \left( {\frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}} \right) \div $$ Â Â $$\left( {\frac{{{x^3} - {y^3} - 3xy\left( {x - y} \right)}}{{x - y}}} \right)$$
$$\eqalign{
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div \frac{{{{\left( {x - y} \right)}^3}}}{{x - y}} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div {\left( {x - y} \right)^2} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \times \frac{1}{{{{\left( {x - y} \right)}^2}}} \cr
& = \frac{1}{{{x^2} + {y^2}}} \cr} $$
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