A) 25
B) 125
C) 625
D) 0
Correct Answer: D
Solution :
\[x+y=15~~~\] \[\Rightarrow (x-10)+(y-5)=0\] \[\therefore {{(x-10)}^{3}}+{{(y-5)}^{3}}\] \[={{\left( x-10+y-5 \right)}^{3}}-3\text{ }\left( x-10 \right)\text{ }(y-5)\] \[(x-10+y-5)=0\] \[[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)]\]You need to login to perform this action.
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