<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] By exploiting $a^3+b^3=(a+b)(a^2-ab+b^2)$ we have:
$$ 21^3+23^3+25^3+27^3 = 48\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$$
hence $ 21^3+23^3+25^3+27^3$ is a multiple of $96$, since $\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$ is an even number.
</s>