(3x - 2y) : (2x + 3y) = 5 : 6, then one of the value of $${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}$$ is?
A. $$\frac{1}{{25}}$$
B. 5
C. $$\frac{1}{5}$$
D. 25
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {2x + 3y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \therefore {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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