how to solve numbers and letters in a braket to the power of somthing

Question

(2y+1)^3

Answer 1

(2y+1)³

One way to solve this is to do the multiplication in stages:

Start by multiplying (2y+1) (2y+1) keeping the result in brackets and combine like terms

= (2y+1) (2y+1)(2y+1)

= (4y²+ 2y + 2y + 1)(2y+1)

Next multiply using the distributive property

= (4y² + 4y +1) (2y + 1)

= (8y³ + 8y² + 2y + 4y² + 4y + 1)

= 8y³ + 12y² + 6y + 1