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P(A∪B) = P(A)P(B)−P(A∩B) Proof There is A∪(B∩Ac) = (A∪B)∩(A∪Ac) = A∪B, which is to say that A∪B can be expressed as the union of two disjoint sets Therefore, according to axiom 3, there is P(A∪B) = P(A)P(B ∩Ac) But B = B ∩ (A ∪ Ac) = (B ∩ A) ∪ (B ∩ Ac) is also the union of two disjoint sets, so there is alsoDon't forget to like, comment, and subscribe!!!Subscribe for new videos wwwyoutubecom/c/MrSalMathShare this video https//youtube/IoqYK6YlyBIThe problemRegister for FREE at http//deltastepcom or download our mobile app https//bitly/3akrBoz to get all learning resources as per ICSE, CBSE, IB, Cambridge &

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A^3+b^3+c^3-3abc proof-`(ab)^3 = a^3 3a^2*b 3ab^2 b^3` You should factor out 3ab in the group `3a^2*b 3ab^2` such that `(ab)^3 = a^3 b^3 3ab(ab)`For all positive integers a and b, LCM(a,b) is unique Pf (We shall omit the proof of the existence of the LCM and just show it's uniqueness, assuming that it exists) Let a and b be positive integers Suppose m 1 and m 2 are two LCM's for a and b Since m 1 is an LCM and m 2

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Identity 3 Let A, B and C be sets Show that (A−B)−C = A−(B ∪C) Proof (A−B)−C = (A∩)−C set difference = (A∩)∩Cc set difference = A∩( ∩Cc) associative = A∩(B ∪C)c de Morgan's = A−(B ∪C) set difference Proof Let x ∈ (A − B) − C Then x ∈ (A − B) and x 6∈C by definition of set differenceA )(B )C) (A and B) )C conditional proof In a course that discusses mathematical logic, one uses truth tables to prove the above tautologies 2 Sets A set is a collection of objects, which are called elements or members of the set Two sets are equal when they have the same elementsProve ABC =(AB)(AC) jaisonshereen asked on Puzzles / Riddles;

Transitive If ∠ ≅∠A B and ,∠ ≅∠B C then ∠ ≅∠A C Notes Writing a TwoColumn Proof In a proof, you make one statement at a time until you reach the conclusion Because you make statements based on facts, you are using deductive reasoning Usually the first statementandreason pair you write is given informationThe expansion of a minus b whole squared algebraic identity can be derived in algebraic form by the geometrical approach The concept of areas of geometrical shapes such as squares and rectangles are used for proving the a minus b whole square formula in algebraic form114 (e) Prove that A∩B and A\B are disjoint, and that A = (A∩B)∪ (A\B) Proof For the first part we have to prove that (A ∩ B) ∩ (A \ B) = ∅ Let x ∈ (A ∩ B) ∩ (A \ B) Then x ∈ A ∩ B and x ∈ A \ B, so x ∈ A and x ∈ B, and x ∈ A and x /∈ B In particular, this implies x ∈ B and x /∈ B, which is a

Summary (AB)^3 If you have any issues in the (AB)^3 formulas, please let me know through social media and mail A Plus B Whole Cube is most important algebra maths formulas for class 6 to 13Proof by induction applied to a geometric series Alisoncom Proof of the Sum of Geometric Series Project Maths Site Further proof by induction – Multiples of 3 Alisoncom Factorisation results such as 3 is a factor of 4n–1 Proj Maths Site 1 Proj Maths SIte 2 Prove De Moivre's Theorem Further proof by induction – Factorials andSince (ab)^3= a^3b^33ab(ab) =>(ab)^33ab(ab)=a^3b^3 =>(ab)((ab)^23ab)=a^3b^3 =>(ab)(a^2b^2ab)=(ab)^3

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Let me help you with this formula in detail (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) Proof (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c)Any integer can be factored into prime numbers, its 'divisors' for example, 60 = 5 x 3 x 2 x 2 The conjecture roughly states that if a lot of small primes divide two numbers a and b , thenFor all integers \(a\) and \(b\text{,}\) if \(a\) and \(b\) are even, then \(ab\) is even There are numbers \(a\) and \(b\) such that \(ab\) is even but \(a\) and \(b\) are not both even False For example, \(a = 3\) and \(b = 5\text{}\) \(ab = 8\text{,}\) but neither \(a\) nor \(b\) are even

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Multiplying the binomial a b by itself three times is the mathematical meaning of cube of the binomial a b So, the a plus b whole cube can be expressed in product form by multiplying three same binomials ( a b) 3 = ( a b) × ( a b) × ( a b) Multiplying three same binomials is a special case in mathematics2 2 Problem 9 of worksheet 3 Proposition 9 For any a;b 2Z, it follows that (a b)3 a3 b3(mod3) Proof Suppose a;b 2Z Thus, there is an integer x such that x = a2b ab2Hence 3x = 3a2b 3ab2 = (a b)3 a3 b3 It follows that 3j(a b)3 3a 3 b3Therefore, (a b) a3 b (mod3) ProofDavneet Singh is a graduate from Indian Institute of Technology, Kanpur He has been teaching from the past 9 years He provides courses for Maths and Science at Teachoo

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Proof (ab)• (a2abb2) = a3a2bab2ba2b2ab3 = a3 (a2bba2) (ab2b2a)b3 = a300b3 = a3b3 Check b3 is the cube of b1 Check a3 is the cube of a1 Factorization is (b a) • (b2 ba a2)So, Area of square of length (ab) = (ab) 2 = (i) (ii) (iii) (iv) Therefore (ab) 2 = a 2 ba ba b 2 ie (ab) 2 = a 2 2ab b 2 Hence Proved This simple formula is also used in proving The Pythagoras Theorem Pythagoras Theorem is one of the first proof in MathematicsThis is basically the proof of 0a=0 hidden within this other proof $\endgroup$ – Sebastian Garrido Dec 14 '13 at 1544 $\begingroup$ @SebastianGarrido that's right I have now colored that part in blue and hopefully this make everyone easier to see $\endgroup$ – achille hui Dec 14 '13 at 1950

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A(BC) = (AB)C Proof If A, B and C are three variables, then the grouping of 3 variables with 2 variables in each set will be of 3 types, such as (A B), (B C) and(C A) According to associative law (A B C) = (A B) C = A (B C) = B (C A) We know that, A AB = A (according to Absorption law)Note the quality q(a, b, c) of the triple (a, b, c) is defined above Refined forms, generalizations and related statements The abc conjecture is an integer analogue of the Mason–Stothers theorem for polynomials A strengthening, proposed by Baker (1998), states that in the abc conjecture one can replace rad(abc) by ε −ω rad(abc), where ω is the total number of distinct primesProof for the Change of Base Rule Proof Step 1 Let x = log a b Step 2 Write in exponent form a x = b Step 3 Take log c of both sides and evaluate log c a x = log c b xlog c a = log c b Videos Proof of the logarithm properties Proof of Product Rule log A log B = log AB Show Stepbystep Solutions

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A )B Equivalence A if and only if B A ,B Here are some examples of conjunction, disjunction and negation x > 1 and x < 3 This is true when x is in the open interval (1;3) x > 1 or x < 3 This is true for all real numbers x(x > 1) This is the same as x 1 Here are two logical statements that are true x > 4 )x > 2 x2 = 1 ,(x = 1 or x = 1)Using algebra and a little deception, Mr John Hush proves (or does he?) that 1 = 2 to a class of amazed calculus studentsIn mathematics, Weitzenböck's inequality, named after Roland Weitzenböck, states that for a triangle of side lengths , , , and area , the following inequality holds ≥ Equality occurs if and only if the triangle is equilateral Pedoe's inequality is a generalization of Weitzenböck's inequality The Hadwiger–Finsler inequality is a strengthened version of Weitzenböck's inequality

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J) a rational number means that it has the form a/b where a,b ∈ Z and b 6= 0 3 A theorem can by symbolized in logical notation as (P ∨ Q) ∧ R ⇒ S ∧ ∼ W Outline the framework you would use (assumptions and goals) for the following You are expected to work any negations through as much as possible a) Direct proof ASSUMPTION (P ∨ Q) ∧ RThus by de nition (a b)3 = a3 3a2b 3ab b3 Proof Let a;b;2Z Then (ab)3 = a3 3a2b3ab2 b 3 Hence (ab)3 (a3 b3) = a 3a2b3ab2 b3 a3 b3 This is equivalent to 3a2b 3ab 2= 3(a2b ab2) We see that a2b ab is an integer, thus by the de nition of divisibility which states that for any integer x;y, we say that x divides y if there is an integer c such that y = xc, 3 j(a b)3 (a3 b3)1 (a b)2 = a2 2ab b2 2 (a b)2 = a2 2ab b2 3 (a b)3 = a3 b3 3ab(a b) 4 (a b)3 = a3 b3 3ab(a b) 5 (a b c)2 = a2 b2 c2 2ab2bc

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Here is animated view of formula Cube of a sum =(ab)³=a³3a²b3ab²b³ You can see both cubes and the six rectangular parallelepipeds in 3Dview Cube of a difference The formula is (ab)³=a³3a²b3ab²b³ You convert it to (ab)³=a³3ab(ab)bLearn how to derive the expansion of a plus b whole cube formula in algebraic approach by the special product of three same binomials mathematicsLemma 33 Let a,b be integers with a > b > 0 Then gcd (a,b) =gcd(b,a−b) Proof Let gcd(a,b) = d and gcd(a−b,b) = e da and db so d(a−b) (by 31) Hence d ≤ e (as e is the greatest common divisor of b and a−b) Also e(a−b) and eb so ea and hence e ≤ d (as d is the greatest common divisor of a and b) Thus d = e

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6 a2 −b2 =(ab)(a−b) 7 a3 −b3 =(a−b)(a2 ab b2) 8 a3 b3 =(ab)(a2 −ab b2) 9 a n−bn=(a−b)(an−1 a −2b an−3b2 bn−1) 10 an= aaantimes 11 ama n= am 12 am an = am−nif m>n =1 ifm=n = 1 an−m if mSo, Area of square of length (ab) = (ab) 2 = (i) (ii) (iii) (iv) Therefore (ab) 2 = a 2 ba ba b 2 ie (ab) 2 = a 2 2ab b 2 Hence Proved This simple formula is also used in proving The Pythagoras Theorem Pythagoras Theorem is one of the first proof in Mathematics2 PROOFS (1) A 1 is invertible with inverse A (2) AB is invertible with inverse B 1A (3) AT is invertible with inverse (A 1)T Proof — Part (1) Since A is invertible, it follows that 9A 1 such that AA 1 = A 1A= I n However, by definition this immediately implies that A 1 is invertible with inverse A Part (2) Since A,B are invertible, it follows that 9A 1;B 1such that AA = A 1A=

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Cube Formulas (a b) 3 = a 3 b 3 3ab (a b) (a − b) 3 = a 3 b 3 3ab (a b) a 3 − b 3 = (a − b) (a 2 b 2 ab) a 3 b 3 = (a b) (a 2 b 2 − ab) (a b c) 3 = a 3 b 3 c 3 3 (a b) (b c) (c a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac) If (a b c) = 0,27 Comments 1 Solution 21,561 Views 1 Endorsement Last Modified prove ABC =(AB)(AC) Comment Premium Content You need a subscription to comment Start Free Trial Watch QuestionBy definition of ( ) symbol (a) means additive inverse of (a) and (b) means additive inverse of (b) As a * 0 = 0 by multiplicative property of zero a * {b (b)} = 0 using additive inverse a*b a(b) = 0 multiplicative associative pro

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Proof Formula \((abc)^3 = \\a^3 b^3 c^3 6abc \\ 3ab (ab) 3ac (ac) 3bc (bc) \) Summary (abc)^3 If you have any issues in the (abc)^3 formulas, please let me know through social media and mailProof Formula \(=> a^3 b^3 = (ab) (a^2 b^2 – ab) \) Verify \( a^3 b^3 \) Formula Need to verify \( a^3 b^3 \) formula is right or wrong put the value of a =2 and b=3 put the value of a and b in the LHS \( => a^3 b^3 = 2^3 3^3 \) \( = > a^3 b^3 = 8 27 = 35 \) put the value of a and b in the RHS \(=> (ab) (a^2 b^2 – ab) \) \(=> (23) (2^2 3^2 2 \times 3) \) \(=> (5) (4 9 – 6) \) \(=> (5) (7) = 35 \) Therefore \( LHS = RHS \) LHR = left hand side, RHS = right hand2 PROOFS (1) A 1 is invertible with inverse A (2) AB is invertible with inverse B 1A (3) AT is invertible with inverse (A 1)T Proof — Part (1) Since A is invertible, it follows that 9A 1 such that AA 1 = A 1A= I n However, by definition this immediately implies that A 1 is invertible with inverse A Part (2) Since A,B are invertible, it follows that 9A 1;B 1such that AA = A 1A=

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In summary, we have shown that if \(A \times C = B \times C\), then \(A = B\) This completes the proof Now we'll look at another way that set operations are similar to operations on numbers From algebra you are familiar with the distributive property \(a \cdot (bc) = a \cdot ba \cdot c\) Replace the numbers a,b,c with sets A, B, C, and= (b^3)(2n^3 3n^2 n)/(6n^3) The area from 0 to a, A_2, can be approximated as a/n • (Σ (ai/n)^2 i = 1 to n) = (a^3)(2n^3 3n^2 n)/(6n^3) Then an approximation of the area from a to b is A_1 A_2 = (b^3 a^3)(2n^3 3n^2 n)/(6n^3) The exact area, then, is lim_{n→∞} (b^3 a^3)(2n^3 3n^2 n)/(6n^3) = 2(b^3 a^3)/6 = (b^3 a^3)/3Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 – 90abc) can be written as (2a) 3 (3b) 3 (5c) 3 – 3 (2a) (3b) (5c) And this represents identity a 3 b 3 c 3 3abc = (a b c) (a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c

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Identity VII (a − b) 3 = a 3 b 3 3ab (a b) Check Algebra Formulas for full list of formulas Lets take an example Identity VI (a b) 3 = a 3 b 3 3ab (a b) (2 3) 3 = 2 3 3 3 3 (2) (3) (2 3) (5) 3 = 8 27 18 × 5 125 = 8 27 90 125 = 125 Identity VII (a − b)3 = a3 b3 3ab (a b)Theorem Let A = {n n = 4k 1 for some k ∈ Z} and B = {n n = 4k −3 for some k ∈ Z} Prove A = B Proof We must show that A ⊆ B and B ⊆ A First, we show that A ⊆ B Let x ∈ A By definition of A, x = 4m1 for some m ∈ Z Letting n = m 1, we check by substitution that 4n − 3 = 4(m 1)− 3 = 4m 4− 3 = 4m 1 = xRecall (Definition 13) that if A and B are sets, then A ⊆ B means that every element of A is also an element of B In other words, it means if a ∈ A, then a ∈ B Therefore to prove that A ⊆ B, we just need to prove that the conditional statement "If a ∈ A, then a ∈ B " is true

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I guess another way of looking at it would be (ab)^3 = a^3 b^3 3(a^2)b 3a(b^2) Remember this formula It is simpler P now, to find the value of a^3 b^3, we move all the other terms to the left which gives us (ab)^3 (3(a^2)b 3a(b= (a b)(a b)(a b) = (a b)(a² ab ab b²) = (a b)(a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³Example Solve 8a 3 27b 3 125c 3 – 90abc Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 – 90abc) can be written as (2a) 3 (3b) 3 (5c) 3 – 3(2a)(3b)(5c) And this represents identity a 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the LHS of identity ie a 3 b 3 c 3

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