A) \[\frac{4x}{3y}\]
B) \[2\left( 4{{x}^{2}}+\frac{1}{9{{y}^{2}}} \right)\]
C) \[\frac{8x}{3y}\]
D) \[\frac{4y}{3x}\]
Correct Answer: C
Solution :
\[{{\left( 2x+\frac{1}{3y} \right)}^{2}}-{{\left( 2x-\frac{1}{3y} \right)}^{2}}\] \[=\left[ 2x+\frac{1}{3y} \right]\left[ 2x+\frac{1}{3y} \right]-\left\{ \left( 2x-\frac{1}{3y} \right)\,\left( 2x-\frac{1}{3y} \right) \right\}\]\[=\left[ 4{{x}^{2}}+\frac{1}{9{{y}^{2}}}+\frac{4x}{3y} \right]-\left\{ 4x+\frac{1}{9{{y}^{2}}}-\frac{4x}{3y} \right\}=\frac{8x}{3y}\]
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