Algebraic calculation, (1)(a+2b-3c)(a-2b+3c) (2) Square of X (X-5) - x (X-5) (X-5) (3) The square of (3x + 1) and the square of (3x-1)

Algebraic calculation, (1)(a+2b-3c)(a-2b+3c) (2) Square of X (X-5) - x (X-5) (X-5) (3) The square of (3x + 1) and the square of (3x-1)

1 original formula = [a + (2b-3c)] [a - (2b-3c)]
=a^2-(2b-3c)^2
=a^2-4b^2+12bc-9c^2
3 original = (3x + 1 + 3x-1) (3x + 1-3x + 1)
=6x *2
=12x
The second question is not very clear

Operation of algebraic expression
Given a & sup2; + A-1 = 0, then a & sup3; + 2A & sup2; + 3 = (), which Einstein did

∵a²+a-1=0 …… ①
Then a (A & sup2; + A-1) = 0
That is a & sup3; + A & sup2; - a = 0 ②
From ① + ②, it can be concluded that:
a³+2a²-1=0
∴a³+2a²+3=4

Algebraic calculation
Kneel down for (2 + 1) × (2 × 2 + 1) × (2 × 2 × 2 + 1) × (2×…… 2 + 1) (n times 2)

（2＋1）×（2×2＋1）×(2×2×2×2＋1)×……… (2×…… 2+1)
The numerator and denominator are all multiplied by (2-1)
Then the original formula = [(2-1) (2 + 1) (2 & sup2; + 1) +... (2 ^ 2n + 1)] / (2-1)
=(2²-1)(2²+1).(2^2n+1)
=2^4n+1
It should be 2n 2 times, right?

Complete square formula
A + B = negative 2, A-B = negative 15, find the square of a - the square of B

a^2-b^2=(a+b)(a-b)=-2X(-15)=30