[最新] (a+b+c)^3 expansion formula 268121-(a+b)^3 expansion formula
(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
The Binomial Theorem Binomial Expansions Using Pascal S Triangle Subsets
I Am Taking A Taylor Series Expansion Of A Function F X I Wish To Expand The Function Around The Value G X Is This Allowable
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