commutativity : for any two literals `a and b` we have `a+b=b+a`
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in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] associative law of addition:
a + (b+c) = (a+b) + c

commutative law of addition:
a + b = b + a

additive inverse:
a + (-a) = 0

zero property:
a + 0 = a

subtraction is basically the same thing as addition.
a - b = a + (-b)
instead of subtracting b, you add its additive inverse. let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, we have to show that: a = b
given: p (a) = p (b)
let x be any element of set a,p(a) is the power set of set a, hence it contains all the subsets of set a. thus set a is contained in set p(a).

now let c be an element of set b
∴ 
now, we have:
c ⊂ b
∴ 
∴ a ⊂ b

similarly, we have:
b ⊂ a
now if a ⊂ b and b ⊂ a
∴ a = b given to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved. you are correct.there is a simple mathematical reason why your answer is correct, but wyzant told me that adding it to this answer required review by wyzant.try again:p(a or b) = p(a) + p(b) - p(a and b)p(neither a nor b) = 1 - p(a or b) let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b 2, 0 ( any two integers with one of them as 0). multiply 2 in both lhs and rhs,integrating on both the sides, we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12