on expanding,
=A^3 - i^3 - 3A^2 + 3A + A^3 + I^3 + 3A^2 + 3A -7A
=2A^3 +6A-7A
=2A^3-A
as we know that A^2=I
A^3=^2 *A=IA=A
the simplified form = 2A-A=A
thus A is the required answer.
0
Priyanka Bishnoi answered this
= a
-2
Archana answered this
how a its coming 8A-7I
2
Aman answered this
Answer is A as after simplying the ques it comes to be 8A -7A
1
Ritu Agarwal answered this
(a) We have, A2 = I
Also A and I are commutative, so we can expand (A+I)n using expansion of (a+b)n, where a and b ∈ C
∴ (A-I)3 + (A+I)3 - 7A
= A3 - 3A2 + 3A - I3 + A3 + 3A2 + 3A + I3 - 7A
= 2A3 + 6A - 7A
= 2A2 . A + 6A - 7A
= 2I.A + 6A - 7A
= 2A + 6A - 7A = 8A - 7A = A
6
+919xxxxxx03 answered this
Apply (a-b)^3 and (a+b)^3 so,
3
Sanika answered this
Answer is A or if you are using I then it is I
1
Surendra Kumar answered this
We have, A2 = I
Also A and I are commutative, so we can expand (A+I)n using expansion of (a+b)n, where a and b ∈ C
∴ (A-I)3 + (A+I)3 - 7A
= A3 - 3A2 + 3A - I3 + A3 + 3A2 + 3A + I3 - 7A
= 2A3 + 6A - 7A
= 2A2 . A + 6A - 7A
= 2I.A + 6A - 7A
= 2A + 6A - 7A = 8A - 7A = A