Answer:
1 + 3sinA + 3sin²A + sin³A
Explanation:
Given
(1 + sinA)(1 + sinA)(1 + sinA) ← expand the last pair using FOIL
= (1 + sinA)(1 + 2sinA + sin²A) ← distribute
= 1 + 2sinA + sin²A + sinA + 2sin²A + sin³A ← collect like terms
= 1 + 3sinA + 3sin²A + sin³A