Unable to solve congruence modulo arithmetic question for power of integer: $(7552)^3 \equiv ... \pmod{89}$.

Question

I am doing simple question on congruence modulo arithmetic for power of a number, and am unable to get correct answer:

$(7552)^3 \equiv ... \pmod{89}$

=> $7552 \equiv (89*(100) - 1348)$
$ \equiv (89*(100) -890 - (1348-890)) \implies (89*(100) -890 -445 - (458 -89*5))$
$\equiv (89*(100) -890 -445 - (458 -89*5)) \equiv (89*(100) -890 -445 - (458 -89*5))$
$\equiv -13$.

So, $(7552) \equiv -13 \pmod{89}$.

So, need find the reduced problem of: $-13^3 \equiv ... \pmod{89}$.

$-169*13 \equiv ... \pmod{89} \implies -80*13 \equiv ... \pmod{89} \implies -1040 \equiv ... \pmod{89}$
$=> -1040 \equiv -(890 + 89 + 61) \implies -61$ should be the answer.

But, the answer is given as $28$.

Answer 1

i would prove that $$7552\equiv 76\mod 89$$