救急 F.2 Identities

If (x-1)^2 -A (x-1) ≡ x^2-B, find the values of A, B and C
L.H.S: (x-1)^2 -A (x-1)
= x^2 -2x +1 -Ax +A
= x^2 -(2 +A)x +(1 +A)
-(2 +A) = 0
A = -2 #
1 +A = -B
B = 1 #

If (x+1)^2 -A (2x-3) ≡ x^2- 2Bx +4B, find the values of A and B
L.H.S: (x+1)^2 -A (2x-3)
= x^2 +2x +1 -2Ax +3A
= x^2 +(2 -2A)x +(3A +1)
2 -2A = -2B
B = A -1 (1)
Put (1) into (3A +1 = 4B)
3A +1 = 4A -4
A = 5 #
B = 4 #

If (Ax+B)^2 ≡ Cx^2 +Dx +E, give 3 sets of the values of A, B, C, D and E which make the identity holds.
L.H.S.: (Ax+B)^2 = (Ax)^2 +2ABx +B^2
(1) A^2 = C
(2) 2AB = D
(3) B^2 = E

Determine whether each of the following equations is an identity
(x-y)(y-z)(z-x) = xyz -x^2 -y^2 -z^2
L.H.S.: (xy -xz -y^2 +yz)(z -x)
= xyz -xz^2 -zy^2 +yz^2 -(x^2)y +(x^2)z +xy^2 -xyz
= -xz^2 -zy^2 +yz^2 -(x^2)y +(x^2)z +xy^2
so this equations is not an identity

Expand (a-b)(a+b)(a^2+b^2)( a^4+b^4)( a^8+b^8)( a^16+b^16)….(a^1024+b^1024). Briefly explain how you get the answer.
(a-b)(a+b)(a^2+b^2)( a^4+b^4)( a^8+b^8)( a^16+b^16)….(a^1024+b^1024)
=(a^2 -b^2)(a^2+b^2)( a^4+b^4)( a^8+b^8)( a^16+b^16)….(a^1024+b^1024)
=( a^4 -b^4)( a^4+b^4)( a^8+b^8)( a^16+b^16)….(a^1024+b^1024)
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=(a^1024 -b^1024)(a^1024+b^1024)
=a^2048 -b^2048