Question:

(a^2 - b^2) > 6 (a^2 - b^2)

This is possible when a^2 - b^2 is negative.

a^2 - b^2 < 0

Is a^2 < b^2 ?

A: b<a consider a=1; b=2 and a=-3; b=2 INSUFFICIENT

B: a<-1 nothing known about B. INSUFFICIENT

Consider a and b,

a = [-.5, 0, 1, 2, 3 .. ] i.e., A can be both negative and positive and b will always be postive.
This is something very similar to what we had in option A.

Thus E.

HSPA wrote:Is the value of a^2 - b^2 greater than the value of (3a + 3b)(2a - 2b) ?

(1) b < a

(2) a < -1

3(a+b)*2(a-b)=6(a^2-b^2);

1) b<a; a>0; b<0; let a=2 b=-3 a^2-b^2=4-9=-5 ; 6(a^2-b^2)=-30;
a>0, b>0 let a=4,b=2 a^2-b^2=16-4=12; 6(a^2-b^2)=6*12=72;

a<0, b<0 a=-1, b=-2 a^2-b^2=1-4=-3, 6(a^2-b^2)=-18;
therefore 1 alone is not sufficient to answer the question.

2) a<-1, as nothing is mentioned about b, therefore 2 alone is not sufficient to answer the question.

consider 1 and 2 together we have, b<a, and a<-1 now consider a=-2, b=-3 a^2-b^2= 4-9=-5; 6(a^2-b^2)= -30, which is less than a^2-b^2 hence C

sorry pranay