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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) 24 Binomial Expansions 41 Pascal's riTangle The expansion of (ax)2 is (ax)2 = a2 2axx2 Hence, (ax)3 = (ax)(ax)2 = (ax)(a2 2axx2) = a3 (12)a 2x(21)ax x 3= a3 3a2x3ax2 x urther,F (ax)4 = (ax)(ax)4 = (ax)(a3 3a2x3ax2 x3) = a4 (13)a3x(33)a2x2 (31)ax3 x4 = a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x)n come from the nth row of Pascal's36 The Binomial and Multinomial Theorems We have previously learned that a binomial is an expression that contains 2 terms and a multinomial is any expression that contains more than 1 term (so a binomial is actually a special case of a multinomial) We have also previously seen how a binomial squared can be expanded using the distributive law For example
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(a+b+c)^3 formula expansion
(a+b+c)^3 formula expansion-0 0 Todd Lv 6So you divide both sides by (abc)3 and you get 9>1 which is always a true statement 0 0 Geetha 6 years ago I just want to ask one question If a b c = 15, then can we show that a^3 b^3 c^3 exactly divisible by 6 or it leaves no remainder on dividing it by 6?
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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) Generally It is, of course, often impractical to write out Pascal"s triangle every time, when all that we need to know are the entries on the nth line= 3*2*1 = 6, 2!A^3 3a^2b 3ab^2 b^3 Use the Binomial expansion (note the exponents sum to the power in each term) (xy)^3 = _3C_0x^3y^0 _3C_1x^2y^1 _3C_2x^1y^2 _3C_3x^0y^3 Remember 3!
(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;= 1 _3C_0 = (3!)/((30)!(0!)) = (3!)/((3)!1) = 1 _3C_1 = (3!)/((31)!(1!)) = (3!)/((2)!1) = (3*2!)/(2!) = 3 _3C_2 = (3!)/((32)!(2!)) = (3!)/((1)!2!) = (3*2!)/(2!) = 3 _3C_3 = (3!)/((33)!(3!)) = (3!)/(0!*3!) = 1 Note (ab)^3 = (a (b))^3 Substitute into the Binomial(a 3 3a 2 b 3ab 2 b 3)(ab) = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4 The calculations get longer and longer as we go, but there is some kind of pattern developing That pattern is summed up by the Binomial Theorem
So you divide both sides by (abc)3 and you get 9>1 which is always a true statement 0 0 Geetha 6 years ago I just want to ask one question If a b c = 15, then can we show that a^3 b^3 c^3 exactly divisible by 6 or it leaves no remainder on dividing it by 6?Free expand & simplify calculator Expand and simplify equations stepbystep3(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 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 )
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2 29 if a ib=0 wherei= p −1, then a= b=0 30 if a ib= x iy,wherei= p −1, then a= xand b= y 31 The roots of the quadratic equationax2bxc=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b p 2a −b− p 2a where = discriminant = b2 −4ac 32Quite unfortunately not from the simple perspective requested, I have found that P = a 3 b 3 c 3 3abc = 2S, where S is the area of the triangle of vertices= 2*1 = 2, 1!
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(abc) 3 a 3 b 3 c 3 We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3 The coefficient of the cubes is therefore 1 (It's the same for a, b and c, of course) 3a 2 b3a 2 c Next, we consider the a 2 terms We can choose two a's from 3 factors in C(3,2) ways=3Note again The lower index, in this case 4, is the exponent of b This same number is also the coefficient of a 4 b 8, since 12 C 8 = 12 C 4 Example 2 Expand (a − b) 5 SolutionWe found the binomial coefficients to be 1 5 10 10 5 1 The difference with (a − b) is that the signs of the terms will alternate(a − b) 5 = a 5 − 5a 4 b 10a 3 b 2 − 10a 2 b 3 5ab 4 − b 5While considering this question;
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Expand (a – b – c d) 10 using multinomial theorem and by using coefficient property we can obtain the required result Using multinomial theorem, we have We want to get coefficient of a 3 b 2 c 4 d this implies that r 1 = 3, r 2 = 2, r 3 = 4, r 4 = 1, ∴ The coefficient of a 3 b 2 c 4 d is (10)!/(3!2!4) (1) 2 (1)4 =Introduction to a plus b whole cube formula with example problems with proofs to learn how to derive ab whole cube identity in mathematics3The slope of the line through them, m = y 2 y 1 x 2 x 1 = rise run Lines can be represented in three di erent ways Standard Form ax by = c SlopeIntercept Form y = mx b PointSlope Form y y 1 = m(x x 1) where a;b;c are real numbers, m is the slope, b (di erent from the standard form b) is the yintercept, and (x 1;y 1) is any xed point on
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A 3 b 3 c 33abc=(ab) 33a 2 b3ab 2 c 33abc =(abc) 3 3(ab) 2 c3(ab)c 2 3ab(abc) =(abc) 3 3(ab)c(abc)3ab(abc)=(abc)(a 2 b 2 c 2 abbcac)A B C 3 Formula Expansion contoh kad ucapan persaraan guru contoh gambar hubungan antara manusia dengan manusia lain contoh hiasan papan kenyataan pejabat contoh carta organisasi syarikat enterprise contoh isi permohonan lanjutan pas lawatan contoh isi borang tuntutan perjalanan contoh jadual harian di rumah contoh format surat berhentiIn elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial According to the theorem, it is possible to expand the polynomial n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positive integer depending on n and b For example, 4 = x 4 4 x 3 y 6 x 2 y 2 4 x y 3 y 4 {\displaystyle ^{4}=x^{4}4x^{3}y6x^{2}y^{2}4xy^{3}y^{4}} The
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The change in temperature is the same for both, the original length is known for both, and the coefficients of linear expansion can be found from Table 122(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;Quadratic Formula For an equation of the form \(ax^2bxc=0\), you can solve for x using the Quadratic Formula $$ x = \frac{b \pm \sqrt{b^24ac}}{2a} $$ Binomial Theorem \((ab)^1= a b\) \((ab)^2=a^22abb^2\) \((ab)^3=a^33a^2b3ab^2b^3\) \((ab)^4=a^44a^3b6a^2b^24ab^3b^4\) Difference of Squares \(a^2b^2=(ab)(ab)\) Rules of Zero
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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 b³2x 2 5x 3 = 0 In this one a=2, b=5 and c=3 x 2 − 3x = 0 This one is a little more tricky Where is a?= 1 and 0!
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As stated in the title, I'm supposed to show that $(abc)^3 = a^3 b^3 c^3 (abc)(abacbc)$ My reasoning $$(a b c)^3 = (a b) c^3 = (a b)^3 3(a b)^2c 3(a b)c^2 c^3 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careersThe Formula is given below (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) Explanation Let us just start with (abc)² = a² b² c²2ab2bc2ca =a² b² c²2(abbcca) now (abc)² (abc)=(a b c)³=a² b² c²2(abbcca)(abc) =a²abc b²abcc²abc 2(abbcca)abc(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, a 3 b 3 c 3 = 3abc
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A 4 – b 4 = (a – b)(a b)(a 2 b 2)A1/3 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 moreBinomial Theorem The Expansion of (a b)n If n is any positive integer, then ( a b) n = a n n C 1 a n − 1 b n C 2 a n − 2 b 2 ⋯ n C m a n − m b m ⋯ b n Where n C m = combination of n objects taken m at a time Some Example of Binomial Expansion ( a b) 2 = a 2 2 a b b 2 ( a b) 3 = a 3 3 a 2 b 3 a b 2 b 3
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A B C 3 Formula Expansion
Thus the expansion for (a b) 6 is (a b) 6 = 1a 6 6a 5 b 15a 4 b 2 a 3 b 3 15a 2 b 4 6ab 5 1b 6 To find an expansion for (a b) 8, we complete two more rows of Pascal's triangle Thus the expansion of is (a b) 8 = a 8 8a 7 b 28a 6 b 2 56a 5 b 3 70a 4 b 4 56a 3 b 5 28a 2 b 6 8ab 7 b 8 We can generalize ourN ∈ N Then the result will be ∑ i = 0 n n C r x n − r y r n C r x n − r y r n C n − 1 x y n − 1 n C n y nExample 3 If a b c = 36 and a 2 b 2 c 2 = 676, then find the value of (ab bc ca) Solution To get the value of (ab bc ac), we can use the formula or expansion of (a b c) 2 Write the formula / expansion for (a b c) 2 (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ac (a b c) 2 = a 2 b 2 c 2 2(ab bc ac)
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Play this game to review Algebra I Determine the values of a, b, and c for the quadratic equation 4x 2 – 8x = 3 Preview this quiz on Quizizz Determine the values of a, b, and c for the quadratic equation 4x2 – 8x = 3Here a = 3x, b = 2y and c = z 3x (2y) (z) 2 = (3x) 2 (2y) 2 (z) 2 2 (3x) (2y) 2 (2y) (z) 2 (z) (3x) = 9x 2 4y 2 z 2 – 12xy 4yz – 6zx 2 Simplify a b c = 25 and ab bc ca = 59 Find the value of a 2 b 2 c 2 Solution According to the question, a b c = 25When the temperature is 22 °C, there is a gap of 10 x 103 m separating their ends No expansion is possible at the other end of either rod At what temperature will the two bars touch?
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Thermal expansion is large for gases, and relatively small, but not negligible, for liquids and solids Linear thermal expansion is Δ L = α L Δ T , where Δ L is the change in length L, Δ T is the change in temperature, and α is the coefficient of linear expansion, which varies slightly with temperatureIn binomial theorem expansion, the binomial expression is most important in an algebraic equation which holds two different terms Such as a b, a3 b3, etc Let's consider;Here a = 3x, b = 2y and c = z 3x (2y) (z) 2 = (3x) 2 (2y) 2 (z) 2 2 (3x) (2y) 2 (2y) (z) 2 (z) (3x) = 9x 2 4y 2 z 2 – 12xy 4yz – 6zx 2 Simplify a b c = 25 and ab bc ca = 59 Find the value of a 2 b 2 c 2 Solution According to the question, a b c = 25
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In Algebra In Algebra putting two things next to each other usually means to multiply So 3 (ab) means to multiply 3 by (ab) Here is an example of expanding, using variables a, b and c instead of numbers And here is another example involving some numbers Notice the "·" between the 3 and 6 to mean multiply, so 3·6 = 18Adding like terms ie (abc) and we get = a 3 b 3 c 3 – 3abc Hence, a 3 b 3 c 3 3abc = (a b c) (a 2 b 2 c 2 ab bc ca) Following are a few applications to this identity= 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 it
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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 are3(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 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 )0 0 Todd Lv 6
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\(=> (abc)^3 = (a^3ab^2ac^2 2a^2b 2abc 2ca^2)\\ (a^2bb^3bc^2 2ab^2 2b^2c 2abc)\\ c \times (a^2b^2c^2 2ab 2bc 2ca) \) \(=> (abc)^3 = (a^3ab^2ac^2 2a^2b 2abc 2ca^2)\\ (a^2bb^3bc^2 2ab^2 2b^2c 2abc)\\ (ca^2 cb^2 c^3 2abc 2bc^2 2c^2a) \) Arrage value according power and similearThe left most is the Pascal triangle The coefficients are multiplied correspondingly by (1,3,3,1), that is, the last line of the Pascal triangle placing vertically You can get the coefficient triangle in the trinomial expansion by finding the product Exercise Expand HintBinomial 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
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In 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},}Well a=1, as we don't usually write "1x 2 " Or we can use the special Quadratic Formula Just plug in the values of a, b and c, and do the calculations We will look at this method in more detail now About the Quadratic FormulaSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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1 decade ago This is just multiplying out and bookkeeping It's a^3 b^3 c^3 plus 3 of each term having one variable and another one squared like ab^2, b^2c, all 6 combinations of those, thenFree math lessons and math homework help from basic math to algebra, geometry and beyond Students, teachers, parents, and everyone can find solutions to their math problems instantlyAnswer (a) To use the binomial series, we need to change the expansion to the form of (1 u) n So we perform the following steps to get it in the required form `sqrt (4x^2)` `= (4x^2)^ (1/2)` `= 4 (1x^2/4)^ (1/2)` `=4^ (1/2) (1x^2/4)^ (1/2)` `=2 (1x^2/4)^ (1/2)`
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