Answer:

(3ab + 2cd)³

To solve (3ab + 2cd)³, we will use the binomial theorem which states:

(a + b)³ = a³ + 3a²b + 3ab² + b³

We can apply this formula by letting a = 3ab and b = 2cd:

(3ab + 2cd)³ = (3ab)³ + 3(3ab)²(2cd) + 3(3ab)(2cd)² + (2cd)³

Simplifying each term:

(27a³b³) + (54a²b²cd) + (36ab³c²d²) + (8c³d³)

Therefore, (3ab + 2cd)³ = 27a³b³ + 54a²b²cd + 36ab³c²d² + 8c³d³.

(4x + 5y)³

To expand the binomial (4x + 5y)³, we can use the formula for the cube of a binomial, which is:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Substituting 4x for a and 5y for b, we get:

(4x + 5y)³ = (4x)³ + 3(4x)²(5y) + 3(4x)(5y)² + (5y)³

Expanding each term, we have:

64x³ + 240x²y + 300xy² + 125y³

Therefore, (4x + 5y)³ = 64x³ + 240x²y + 300xy² + 125y³.

Step-by-step explanation:

[tex]\purple{Answer}[/tex]

To solve (3ab + 2cd)³, we can use the formula for the cube of a binomial:

(a + b)³ = a³ + 3a²b + 3ab² + b³

In this case, our binomial is (3ab + 2cd), so we have:

(3ab + 2cd)³ = (3ab)³ + 3(3ab)²(2cd) + 3(3ab)(2cd)² + (2cd)³

= 27a³b³ + 54a²b²cd + 36ab(c²d²) + 8c³d³

Therefore, (3ab + 2cd)³ = 27a³b³ + 54a²b²cd + 36abc²d² + 8c³d³.

To solve (4x + 5y)³, we can again use the formula for the cube of a binomial:

(a + b)³ = a³ + 3a²b + 3ab² + b³

In this case, our binomial is (4x + 5y), so we have:

(4x + 5y)³ = (4x)³ + 3(4x)²(5y) + 3(4x)(5y)² + (5y)³

= 64x³ + 240x²y + 300xy² + 125y³

Therefore, (4x + 5y)³ = 64x³ + 240x²y + 300xy² + 125y³.