Answer:
The answer to your question is (5m50−11n8)(5m50+11n8)
Explanation:
25m¹⁰⁰−121n¹⁶
find the prime factors of 25 and 100
25 5 121 11
5 5 11 11
1 1
25 = 5 x 5 = 5² 121 = 11 x 11 = 11²
m¹⁰⁰ = (m⁵⁰)² n¹⁶ = (n⁸)²
Substitution
25m¹⁰⁰−121n¹⁶ = 5²(m⁵⁰)² - 11² (n⁸)²
= (5m⁵⁰ - 11n⁸) (5m⁵⁰ + 11n⁸)
(5m50−11n8)(5m50+11n8)
(5m50−11n8)2
(5m10−11n4)2
(5m10−11n4)(5m10+11n4)