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(a b) 1 = a b (a b) 2 = (a b) * (a b) = a 2 2ab b 2;Explanation Binomial formula for (a b)3 ⇒3 C0a3b0 3 C1a2b1 3 C2a1b2 3 C3a0b3 Here, a = x and b = 1 ⇒3 C0x3 3 C1x2 × 11 3 C2x1 ×12 3 C3 × 13 As →3 C0 =3 C3 = 1 and →3 C1 =3 C2 = 3 ⇒ x3 3x2 3x 1In mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomials The expansion is given by The expansion is given by ( a b c ) n = ∑ i j k = n i , j , k ( n i , j , k ) a i b j c k , {\displaystyle (abc)^{n}=\sum _{\stackrel {i,j,k}{ijk=n}}{n \choose i,j,k}\,a^{i}\,b^{\;\!j}\;\!c^{k},}
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(a+b)^3 expansion formula
(a+b)^3 expansion formula-= a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x) n come from the nth row of Pascal's riangleT , in which each term is the sum of the two terms just above itWhat I want to do with this video is cover something called the triple product expansion or Lagrange's formula, sometimes And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c
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A 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac) If (a b c) = 0, a 3 b 3 c 3 = 3abcThe Formula is given below (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c)= (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 b³ Answer a³ 3a²b 3ab² b³= (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 b³
$$(a b c)^3 = (a^3 b^3 c^3) (3a^2b 3a^2c 3abc) (3ab^2 3b^2c 3abc) (3ac^2 3bc^2 3abc) 3abc$$ $$(a b c)^3 = (a^3 b^3 c^3) 3a(ab ac bc) 3b(ab bc ac) 3c(ac bc ab) 3abc$$3 (abc)2=a2b2c22(abbcca) 4 (ab)3=a3b33ab(ab);a3b3=(ab)−3ab(ab) 5 (a−b)3=a3−b3−3ab(a−b);a3−b3=(a−b)33ab(a−b) 6a2−b2=(ab)(a−b) 7a3−b3=(a−b)(a2abb2) 8a3b3=(ab)(a2−abb2) 9an−bn=(a−b)(an−1a−2ban−3b2 bn−1) 10an=aaantimes 11aman=am 12A^3 b^3 c^3 3abc = (a b c) (a^2 b^2 c^2 ab ac bc) By assumption a^3b^3c^3=3abc so the left hand side is 0 Therefore (abc) (a^2b^2c^2abacbc) = 0 So either abc=0 or
(say, a;b) to functions in the new variable z, whose domain is an interval of 2ˇlength This is because x= a)z= ˇ l aand x= b)z= ˇ l b= 2 b a (b a a) = 2ˇ ˇ l a Thus when the variable xin f(x) moves from ato b, the new variable zin the new function F(z) (which is the same function fin the new variable) moves from cto c2ˇ, where c= ˇ l a\(=> (abc)^3 = (a^3ab^2ac^2 2a^2b 2abc 2ca^2)\\ (a^2bb^3bc^2 2ab^2 2b^2c 2abc)\\ c \times (a^2b^2c^2 2ab 2bc 2ca) \) \(=> (abc)^3 = (a^3ab^2ac^2 2a^2b 2abc 2ca^2)\\ (a^2bb^3bc^2 2ab^2 2b^2c 2abc)\\ (ca^2 cb^2 c^3 2abc 2bc^2 2c^2a) \)The cube of sum of the terms a and b or a binomial is written in the following mathematical form in mathematics ( a b) 3 The a plus b whole cube is equal to the a cubed plus b cubed plus three times the product of a, b and sum of a plus b ( a b) 3 = a 3 b 3 3 a b ( a b)
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(a b) 3 = a 3 b 3 3ab(a b) (a – b) 3 = a 3 – 3a 2 b 3ab 2 – b 3 = a 3 – b 3 – 3ab(a – b) a 3 – b 3 = (a – b)(a 2 ab b 2) a 3 b 3 = (a b)(a 2 – ab b 2) (a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4 (a – b) 4 = a 4 – 4a 3 b 6a 2 b 2 – 4ab 3 b 4;So the answer is 3 3 3 × (3 2 × x) 3 × (x 2 × 3) x 3 (we are replacing a by 3 and b by x in the expansion of (a b) 3 above)B) (a − b) 3 = a 3 − 3a 2 b 3ab 2 − b 3 c) ( x y ) 4 = x 4 4 x 3 y 6 x 2 y 2 4 xy 3 y 4 d) ( x − y ) 4 = x 4 − 4 x 3 y 6 x 2 y 2 − 4 xy 3 y 4
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According to this theorem, it is possible to expand the polynomial \((x y)^n\) into a series of the sum involving terms of the form a \(x^b y^c\) Here the exponents b and c are nonnegative integers with condition that b c = n Also, the coefficient of each term is a specific positive integer depending on n and b For example (for n = 4), we have \((xy)^4=x^44x^3y6x^2y^24xy^3y^4\) It is obvious that such expressions and their expansions would be very painful to multiply out by handThe square of the sum of three or more terms can be determined by the formula of the determination of the square of sum of two terms Now we will learn to expand the square of a trinomial (a b c) Let (b c) = x Then (a b c) 2 = (a x) 2 = a 2 2ax x 2Start with a2 b2 = c2 Put in what we know 12 12 = c2 Calculate squares 1 1 = c2 11=2 2 = c2 Swap sides c2 = 2 Square root of both sides c = √2 Which is about c = It works the other way around, too when the three sides of a triangle make a2 b2 = c2, then the triangle is right angled
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A3 3a2b 3ab2 b3 Now, notice the exponents of a They start at 3 and go down 3, 2, 1, 0 Likewise the exponents of b go upwards 0, 1, 2, 3 If we number the terms 0 to n, we get this k=0 k=1 k=2 k=3Ex a b, a 3 b 3, etc Binomial Theorem Let n ∈ N,x,y,∈ R then n Σ r=0 nC r x n – r · y r nC r x n – r · y r nC n1 x · y n – 1 nC n · y n ie(x y) n = n Σ r=0 nC r x n – r · y r where, Illustration 1 Expand (x/3 2/y) 4 Sol Illustration 2 (√2 1) 5 (√2 − 1) 5 Sol We have5(a b c)2 = a2 b2 c2 2ab2bc 2ca 6(a b c)3 = a3 b3 c33a2b3a2c 3b2c 3b2a 3c2a 3c2a6abc 7a2 b2 = (a b)(a – b ) 8a3 – b3 = (a – b) (a 2 ab b2 ) 9a3 b3 = (a b) (a 2 ab b2 ) 10(a b)2 (a b) 2 = 4ab 11(a b)2 (a b) 2 = 2(a 2 b2 ) 12If a b c =0, then a3 b3 c3 = 3 abc INDICES AND SURDS 1 am a n = a m n 2
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$$(a b c)^3 = (a^3 b^3 c^3) (3a^2b 3a^2c 3abc) (3ab^2 3b^2c 3abc) (3ac^2 3bc^2 3abc) 3abc$$ $$(a b c)^3 = (a^3 b^3 c^3) 3a(ab ac bc) 3b(ab bc ac) 3c(ac bc ab) 3abc$$A 4 – b 4 = (a – b)(a b)(a 2 b 2)`(ab)^n=` `a^nna^(n1)b` `(n(n1))/(2!)a^(n2)b^2` `(n(n1)(n2))/(3!)a^(n3)b^3` `b^n` This can be written more simply as ( a b ) n = n C 0 a n n C 1 a n −1 b n C 2 a n −2 b 2 n C 3 a n −3 b 3 n C n b n
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Popular Answers (1) 5th Feb, 16 Ioulia N Baoulina a 3 b 3 c 3 3abc= (ab) 3 3a 2 b3ab 2 c 3 3abc = (abc) 3 3 (ab) 2 c3 (ab)c 2 3ab (abc) = (abc) 3 3 (ab)c (abc)3ab (aWhat Is The Expansion Of A B C 3 Quora For more information and source, see on this link https//wwwquoracom/Whatistheexpansionofabc3Thermal expansion is large for gases, and relatively small, but not negligible, for liquids and solids Linear thermal expansion is ΔL = αLΔT, where ΔL is the change in length L, ΔT is the change in temperature, and α is the coefficient of linear expansion, which varies slightly with temperature
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What I want to do with this video is cover something called the triple product expansion or Lagrange's formula, sometimes And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross cA 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 3abc = (a b c) (a 2 b 2 c 2 ab bc ca) and we getThe scalar triple product is unchanged under a circular shift of its three operands (a, b, c) ⋅ (×) = ⋅ (×) = ⋅ (×)
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= (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 b³ Answer a³ 3a²b 3ab² b³= (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 b³A³ a²b a²c 2a²b 2ab² 2abc 2a²c 2abc 2ac² ab² b³ b²c 2abc 2b²c 2bc² ac² bc² c³ This is just multiplying out and bookkeeping It's a^3 b^3 c^3 plus 3 of eachAdding like terms, the final formula (worth remembering) is (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ac Practice Exercise for Algebra Module on Expansion of (a b c) 2
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5(a b c)2 = a2 b2 c2 2ab2bc 2ca 6(a b c)3 = a3 b3 c33a2b3a2c 3b2c 3b2a 3c2a 3c2a6abc 7a2 b2 = (a b)(a – b ) 8a3 – b3 = (a – b) (a 2 ab b2 ) 9a3 b3 = (a b) (a 2 ab b2 ) 10(a b)2 (a b) 2 = 4ab 11(a b)2 (a b) 2 = 2(a 2 b2 ) 12If a b c =0, then a3 b3 c3 = 3 abc INDICES AND SURDS 1 am a n = a m n 2So the answer is 3 3 3 × (3 2 × x) 3 × (x 2 × 3) x 3 (we are replacing a by 3 and b by x in the expansion of (a b) 3 above)A³ b³ = (a b)(a² – ab b²) you know that (a b)³ = a³ 3ab(a b) b³ then a³ b³ = (a b)³ – 3ab(a b) = (a b)(a b)² – 3ab = (a b)(a² 2ab b² – 3ab) = (a b)(a² – ab b² ) Please log inor registerto add a comment Related questions
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Adding like terms, the final formula (worth remembering) is (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ac Practice Exercise for Algebra Module on Expansion of (a b c) 2A3 3a2b 3ab2 b3 Now, notice the exponents of a They start at 3 and go down 3, 2, 1, 0 Likewise the exponents of b go upwards 0, 1, 2, 3 If we number the terms 0 to n, we get this k=0 k=1 k=2 k=3(a b) 3 = a 3 b 3 3ab(a b) (a – b) 3 = a 3 – 3a 2 b 3ab 2 – b 3 = a 3 – b 3 – 3ab(a – b) a 3 – b 3 = (a – b)(a 2 ab b 2) a 3 b 3 = (a b)(a 2 – ab b 2) (a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4 (a – b) 4 = a 4 – 4a 3 b 6a 2 b 2 – 4ab 3 b 4;
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(b)If exactly one member must be a woman, there are C(3;1)C(16;3) = 3560 = 1680 ways to form the committee (c)All of the committees formed in (b) qualify except those in which Smith and Jones are both members There are C(3;1)C(14;1) = 42 committees that both Smith and Jones can serve on, so C(3;1)C(16;3) C(3;1)C(14;1) = 1680 42 = 1638 committees canAccording to this theorem, it is possible to expand the polynomial \((x y)^n\) into a series of the sum involving terms of the form a \(x^b y^c\) Here the exponents b and c are nonnegative integers with condition that b c = n Also, the coefficient of each term is a specific positive integer depending on n and b For example (for n = 4), we have \((xy)^4=x^44x^3y6x^2y^24xy^3y^4\) It is obvious that such expressions and their expansions would be very painful to multiply out by handThermal expansion is large for gases, and relatively small, but not negligible, for liquids and solids Linear thermal expansion is ΔL = αLΔT, where ΔL is the change in length L, ΔT is the change in temperature, and α is the coefficient of linear expansion, which varies slightly with temperature
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Take a = 2 x and b = 5 y, and replace them in the expansion of the formula for calculating the value in mathematical form ( 2 x − 5 y) 2 = ( 2 x) 2 ( 5 y) 2 − 2 ( 2 x) ( 5 y) ( 2 x − 5 y) 2 = 4 x 2 25 y 2 − 2 × 2 x × 5 y ∴ ( 2 x − 5 y) 2 = 4 x 2 25 y 2 − x y ( 2) Simplify 9 m 2 16 n 2 − 24 m n(a – b – c) 2 = a 2 b 2 c 2 – 2ab 2bc – 2ca (a b) 3 = a 3 3a 2 b 3ab 2 b 3;Property is if a b c = 0 then a 3 b 3 c 3 = 3abc (i) a = 13, b = 8 and c = 5 13 3 (8) 3 (5) 3 = 3(13)(8)(5) = 1560 (ii) a = 7, b = 3, c = 10 7 3 3 3 (10) 3 = 3(7)(3)(10) = 630 (iii)a = 9, b = 5, c = 4 9 3 5 3 4 3 = 9 3 (5) 3 (4) 3 = 3(9)(5)(4) = 540 (iv) a = 38, b = 26, c = 12 38 3 (26) 3 (12) 3 = 3(38)(26)(12) =
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A 4 – b 4 = (a – b)(a b)(a 2 b 2)Ex a b, a 3 b 3, etc Binomial Theorem Let n ∈ N,x,y,∈ R then n Σ r=0 nC r x n – r · y r nC r x n – r · y r nC n1 x · y n – 1 nC n · y n ie(x y) n = n Σ r=0 nC r x n – r · y r where, Illustration 1 Expand (x/3 2/y) 4 Sol Illustration 2 (√2 1) 5 (√2 − 1) 5 Sol We have(a – b – c) 2 = a 2 b 2 c 2 – 2ab 2bc – 2ca (a b) 3 = a 3 3a 2 b 3ab 2 b 3;
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We can choose an a in 3 ways, and then a b in 2 ways, and then we have only one way to choose a c The coefficient is therefore 3·2·1=6 Finally, add up the coefficients to check they come to (111) 3 =27Solution We have (a b) n, where a = 2/x, b = 3√ x, and n = 4 Then using the binomial theorem, we have Then using the binomial theorem, we have Finally (2/x 3√ x ) 4 = 16/x 4 96/x 5/2 216/x 216x 1/2 81x 2Once you get above the fourth power, the algebra becomes tedious You don't have to calculate these out completely though there's a shortcut of sorts The formula gives the expansion of any binomial series, but you'll still have to work through some algebra to actually expand it
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I could never remember the formula for the Binomial Theorem, so instead, I just learned how it worked I noticed that the powers on each term in the expansion always added up to whatever n was, and that the terms counted up from zero to nReturning to our intial example of (3x – 2) 10, the powers on every term of the expansion will add up to 10, and the powers on the terms will increment byA 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 3abc = (a b c) (a 2 b 2 c 2 ab bc ca) and we getA1/3 a1/3 a1/3 = a (24) (a1/3)3 = a (25) (a2)1/3 = (a1/3)2 = a2∕3 (26) (a1/3)1/4 = a1/3 1/4 = (a1/4)1/3 (27) (a b)1/3 = a1/3 b1/3 (28) (a / b)1/3 = a1/3 / b1/3 (29) (1 / a)1/3 = 1 / a1/3 = a1/3 (30) Sponsored Links Mathematics Mathematical rules and laws numbers, areas, volumes, exponents, trigonometric functions and more
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