Solve (W^2-16/W)(3/4-w) please?

Question is unclear.

Could it be that you mean:-

[ (w² - 16) / w ] [ (3 / (4 - w) ] ??????

[ (w - 4)(w + 4) / w ] [ 3 / (4 - w) ]

[ - (4 - w)(w + 4) / w ] [ 3 / (4 - w) ]

[ - (w + 4) / w ] 3

- 3 (w + 4) / w

Brackets MUST be used !!!

[(w^2 - 16)/w][3/(4 - w)]
= [(w + 4)(w - 4)/w][3/(4 - w)]
= [-1(w + 4)(4 - w)/w][3/(4 - w)] (cancel out 4 - w)
= -3(w + 4)/w
= (-3w - 12)/w
= -3w/w - 12/w
= -3 - 12/w

(w^2-16w^-1)(3/4-w)
(3/4)w^2-w^3-12/w+16