Question

Разложите на множители
64-(3b+1)³

Answer 1

$(4 - (3b + 1)) \times (16 + 4(3b + 1) + (3b + 1) ^{2} ) = (4 - 3b - 1) \times (16 + 12b + 4 + 9b ^{2} + 6b + 1) = (3 - 3b) \times (21 + 18b + 9b ^{2} ) = 3 \times (1 - b) \times 3(7 + 6b + 3b ^{2} ) = 9 \times (1 - b) \times (7 + 6b + 3b ^{2} )$