<s>[INST] Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$? 
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?

MyApproach
Since I cannot form pattern above 
I simplified this as $21^3$ = $3^3$ $. $ $7^3$ 
Similarly I did  $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)=$9$ $.$ $1072$/ $2^5$.
From solving this I get Remainder as $1$
And Similarly on solving $25^3$+$23^3$ separately on dividing by $96$.I get remainder as $73$ and $71$
On adding and Finding remainder I get Remainder as $49$.
I am getting wrong Ans.

Please correct me how to approach towards the problem.

 [/INST] This is essentially a simpler version of lab's answer: 
For any $a,b$, $(a-b)^3+(a+b)^3 = 2a^3 + 6ab^2 \equiv 0 \mod a$ for all $a,b$. In you case, you have two such pairs with $a=24$ (and $b=1,3$ respectively). So the answer is $0$.
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