Question

Can anyone please help me with 1,2,&,3 ?

Answer 1

(x+3)^3 = x^3+9x^2+27x+27

(x+2)(x-2)(x^2+4) = x^4-16

(x-2)^2+1 = x^2-4x+5

Answer 2

Okay, lets start with number 1.

$(x+3)^{3}$ is also equal to (x+3)(x+3)(x+3)

If you take the first two ones and multiple them out it ends up with $x^{2} + 3x +3x + 9$ or also know as $x^{2} + 6x + 9$

You see to multiply them you take the first number in the first group of parenthesis and use the distributive property on the second equation, then you do the same this but with the second number in the first equation.

So once you have $(x^{2} + 6x + 9)$ you multiply it with your remaining (x + 3) using the distributive property to get $x^3 + 9x^2 + 27x + 27$

I hope this helps you get the answers for the other two.