A) 13 : 14
B) 14: 13
C) 10: 11
D) 1: 2
Correct Answer: A
Solution :
\[\because \] \[x:y=3:1\] \[\therefore \] \[\frac{{{x}^{3}}-{{y}^{3}}}{{{x}^{3}}+{{y}^{3}}}=\frac{{{\left( \begin{matrix} x \\ y \\ \end{matrix} \right)}^{3}}-1}{{{\left( \begin{matrix} x \\ y \\ \end{matrix} \right)}^{3}}+1}=\frac{{{(3)}^{3}}-1}{{{(3)}^{3}}+1}\] \[=\frac{27-1}{27+1}\] \[=\frac{26}{28}=13:14\]You need to login to perform this action.
You will be redirected in
3 sec