Vectors A and B are inclined at an angle 120⁰. If |A|=1 , |B|=2, then [(A+3B)x(3A-B)]² = ??

It is equal to [10BXA]²=100(1)²(2)²sin²120⁰=400(3/4)=300

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

= (3(AxA) - AxB + 9(BxA) - 3(BxB))^2

=  (0 - AxB - 9(AxB) - 0 )^2

= (-10(AxB))^2

= (-10(1)(2)(sqrt3/2))^2

= 300

Joy francis gave u a detailed explanation.That's the method I too followed.I hope u'r doubt is clear now.