(b^2 - 3)^3-(b^2 - 3)*(b^4 + 3b^2 + 9)

Question

(b^2 - 3)^3-(b^2 - 3)*(b^4 + 3b^2 + 9)

Answer 1

(b^2-3)^3-(b^2-3)(b^4+3b^2+9)=(b^2-3)^3-(b^6-27)=b^6-9b^4+27b^2-27-b^6+27=27b^2-9b^4=9b^2(3-b^2)