Guys me to solve these questions

Step-by-step explanation:

The first one I would factor the difference of cubes on top and then cancel a common factor this would end up giving you sin^2(A/2)+sin(A/2)cos(A/2)+cos^2(A/2)

(the 1st+3rd term here actually equals 1)

now you have

1+sin(A/2)cos(A/2)

1+1/2*2sin(A/2)cos(A/2)    (this step was just to help show how I'm going to use the next identity)

1+1/2sin(A)                    (double angle identity for sine)

hint for the next one: I would multiply top and bottom by bottom's conjugate first

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option