If log _{10}left(frac{x^{3}-y^{3}}{x^{3}+y^{3 | vedclass

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(C) Given $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$.By the definition of logarithm,$\frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}=100$.There

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$\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$

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(C) Given $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$.By the definition of logarithm,$\frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}=100$.Therefore,$x^{3}-y^{3}=100(x^{3}+y^{3})$.$x^{3}-y^{3}=100x^{3}+100y^{3}$.$-99x^{3}=101y^{3}$.Differentiating both sides with respect to $y$:$-99 \cdot 3x^{2} \frac{dx}{dy} = 101 \cdot 3y^{2}$.$-297x^{2} \frac{dx}{dy} = 303y^{2}$.$\frac{dx}{dy} = -\frac{303y^{2}}{297x^{2}} = -\frac{101y^{2}}{99x^{2}}$.Thus,$\frac{dx}{dy} = \left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$,then $\frac{dx}{dy} = $

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