【ベストコレクション】 (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3 308762

Factor (a^2b^2)^3(b^2c^2)^3(c^2a^2)^3 Use the Binomial Theorem Simplify each term Tap for more steps Multiply the exponents in Tap for more steps Apply the power rule and multiply exponents, Multiply by Rewrite using the commutative property of multiplication Multiply by7a2 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 am m n a an = − 3 (a ) am n mn= 4 (ab) a bm m m= 5 a am m b bmFor some m and n Thus, $E = (a b)(b c)(c a)m(a^2 b^2 c^2) n(ab bc ca)$

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(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3

(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3-A B = 2 3 => A / B = 2 / 3 =(2*4) / (3*4) = 8 / 12 MULTIPLYING numerator & denominator by 4 so, B / A = 12 / 8 ie we can write, B A = 12 8 (I) B C = 4 5 => B / C = 4 / 5 = (4*3) / (5*3) = 12/ 15 multiplying numerator & denominator by 3Get an answer for 'prove that (a^2b^2)^3(b^2c^2)^3(c^2a^2)^3=3(ab)(bc)(ca)(ab)(bc)(ca) no' and find homework help for other Math questions at eNotes We've discounted annual

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Using properties of determinants, show the following ((bc)2,ab,ca),(ab,(ac)2,bc),(ac,bc,(ab)2)=2abc(abc)3 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries(a)(3), is act July 1, 1944, ch 373, 58 Stat 6, which is classified generally to chapter 6A (§ 1 et seq) of Title 42, The Public Health and Welfare For complete classification of this Act to the Code, see Short Title note set out under section 1 of Title 42 and TablesA^3 b^3 = sixty one (a b)(a^2 ab b^2) = sixty one if (a b) = one million, then a^2 ab b^2 = sixty one a b = one million a = b one million a^2 ab b^2 = sixty one (b one million)^2 (b one million)b b^2 = sixty one ((b one million)(b one million)) b^2 b b^2 = sixty one (b^2 b b one million) 2b^2 b = sixty one b^2 2b one million 2b^2 b = sixty one 3b^2 3b one million = sixty one 3b^2 3b 60 = 0 3(b^2 b ) = 0 3(b 5)(b 4

It is called "Pythagoras' Theorem" and can be written in one short equation a 2 b 2 = c 2 Note c is the longest side of the triangle;Views around the world You can reuse this answer Creative Commons LicenseNên (a b c) 2 < a 2 b 2 c 2 (a 2 b 2) (a 2 c 2) (b 2 c 2) = 3(a 2 b 2 c 2) bởi Ngọc Hân 07/11/18 Like (0) Báo cáo sai phạm Cách tích điểm HP Nếu bạn hỏi, bạn chỉ thu về một câu trả lời

1)a b c^2 = a^2 b^2 c^2 2*ab bc ca 2)a^3 b^3 c^3 3abc = (abc)(a^2 b^2 c^2 ab bc ca) Using 1st note 1^2 = 2 2*ab bc ca so ab bc ca = 1/2 (Memorize this as Equation 2) Using 2nd formula in NOTEBy using above equation let us consider a = a, b= b and c = c Then the equation (10 BECOMES a 3 (b 3) (c 3) − 3a (b) (c) = (a b c) (a 2 (b 2) (c 2) − a (b) − (b) (c) − (c)a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 ab – bc ca)(a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca (a (−b) (−c)) 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 Some not so Common Formulas Power n Formula a n − b n

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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 32$$ \begin{align*} \sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\ abc &\geq abc \end{align*} $$ Which is true, but it would imply that equality always holds, which is obviously false So why can't I just divide in a cycling sum?(abcd) 2 a 2 b 2 c 2 d 2 With 2 factors, n=2, we can choose two "a" in one way one from each factor, so the coefficient of the a 2 term is 1 Similarly for the rest of the squared terms Alternatively, we have C(2,2)=1 2ab For the terms containing ab, we can choose an "a" in 2 ways (one from the first factor, or one from the second

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$\frac{a^3}{a^2abb^2}\frac{b^3}{b^2bcc^2}\frac{c^3}{c^2caa^2}\geq \frac{abc}{3}$ posted in Bất đẳng thức và cực trị Cho a, b, c là các số thực dương thỏa mãn $\frac{a^3}{a^2abb^2}\frac{b^3}{b^2bcc^2}\frac{c^3}{c^2caa^2}\geq \frac{abc}{3}$ Để giải bài này ta chứng minh $\frac{a^3}{a^2abb^2}\geq \frac{2ab}{3}$ (Tương đương $(ab)(aIf a,b,c are all nonzero and a b c = 0, prove that a2/bc b2/ca c2/ab = 3 asked Sep 14, 18 in Class IX Maths by aditya23 ( 2,139 points) polynomialsFree PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystep

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The answer is 2^3*3^2*5^1*103*107 as this is a multiple of 12, Now 1284 and 2472 are multiples of that number So HCF of these three is HCF of 1284 and 2472 which is 12 Verify using def GCD(a,b)2g=12(32g1)11 One solution was found g = 1/3 = 0003 Rearrange Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equationGroup 1 (bc 3 1) • (a 2 b 2) Group 2 (b c) • (a 3 b 2 c 2) Group 3 (a 2 b 2) • (c 2) Looking for common subexpressions Group 1 (bc 3 1) • (a 2 b 2) Group 3 (a 2 b 2) • (c 2) Group 2 (b c) • (a 3 b 2 c 2) Bad news !!

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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 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) and we getIf A (2, 1), B (2, 3) and C (2, 4) are three points, then the angle between BA and BC is (a) tan1 (3/2) (b) tan1 (2/3) (c) tan1 (7/4) (d) None of these straight lines;The answer is 2^3*3^2*5^1*103*107 as this is a multiple of 12, Now 1284 and 2472 are multiples of that number So HCF of these three is HCF of 1284 and 2472 which is 12 Verify using def GCD(a,b)

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$$ 2(a^2b^2c^2)^2 5(a^2b^2c^2)(abc)^2 30 abc (abc) 3 (abc)^4 \ge 0 $$ Expanding gives $$ 2 \sum_{cyc} a^4 \sum_{cyc} a^3(bc) 2 \sum_{cyc} a^2 b^2 2 a b c \sum_{cyc} a \ge 0 $$ but I didn't manage to advance furtherSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreEx 14, 4 If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10};

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여기까지가 중학교 과정이다 15 개정 교육과정 기준으로 중학교 3학년 1학기 과정에 나온다 2 09 개정 교육과정에서는 2학년 1학기 과정에서 이 공식들을 배웠으나, 이번에 인수분해와 함께 연계시키기 위해서 3학년으로 올린 듯 3 이 아래부터는 고등 수학 (상)에서 배우게 된다P = (a b)(b c)(c a) is cyclic of degree 3 If E = pq, then q should be homogeneous and cyclic in a, b and c, and is of degree 2 The most general expression of q is then $m(a^2 b^2 c^2) n(ab bc ca)$;Find (i) A ∪ B A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6

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Factoring by pulling out fails The groups have no common factor and can not be added up to form a multiplication Final\(\frac{(a^2b^2)^3(b^2c^2)^3 (c^2a^2)^3}{(ab)^3(bc)^3(ca)^3}\) on simplification is equal to a) (c) (a b) (b c) (c a) (d) 0 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queriesLet xy=a,xy=b Then, we have a^2b=a, ie b=a^2a By the way, since x,y are the real roots of t^2atb=0, we have (a)^24\cdot 1\cdot b\ge 0, ie a^24(a^2a)\ge 0, ie 0\le a\le\frac 43

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Factoring by pulling out fails The groups have no common factor and can not be added up toThe answer is C 3/2 you have abc=3, a^2b^2c^2=6 and 1/a1/b1/c=1=> (abbcca)/abc=1=>abbcca=abc now (abc)^2=a^2b^2c^22ab2bc2ca=>3^2=62 (abbcca) =>9=62abc=>2abc=96=>2abc=3=>abc=3/2 (since abbcca=abc) Show more Source (s) self exmaths teacherA = 2 3 x 3 b = 2 x 3 x 5 c = 3 n x 5 LCM of a,b,c = 2 3 x 3 n x 5 Since, LCM will consists of the highest power of the prime factors So, the power of 3 in finding out the LCM will be n as both a and b have consists of only one 3 as one of their prime factors

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That means we can use this formula here because abc is coming out to be zero (a 2 b 2 ) 3 (b 2 c 2 ) 3 (c 2 a 2 ) 3 = 3(a 2 b 2 )(b 2 c 2 )(c 2 a 2 ) Now let's solve the DenominatorThe formula is (xy)^3 = x^3 3x^2y 3xy^2 y^3 Expand (ab)^3 = a^3 3a^2b 3ab^2 b^3 Expand (bc)^3 = b^3 3b^2c 3bc^2 c^3 Expand (ca)^3 = c^3 3c^2a 3ca^2 a^3 = c^3The formula is (xy)^3 = x^3 3x^2y 3xy^2 y^3 Expand (ab)^3 = a^3 3a^2b 3ab^2 b^3 Expand (bc)^3 = b^3 3b^2c 3bc^2 c^3 Expand (ca)^3 = c^3 3c^2a 3ca^2 a^3 = c^3

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By using above equation let us consider a = a, b= b and c = c Then the equation (10 BECOMES a 3 (b 3) (c 3) − 3a (b) (c) = (a b c) (a 2 (b 2) (c 2) − a (b) − (b) (c) − (c)a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 ab – bc ca)Equations Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations a^3(b^2c^2)b^3(c^2a^2)c^3(a^2b^2) so that you understand better Home3 a) truth table b) sop y0 = (a'b'c'd)(a'b'cd')(a'bc'd')(a'bcd)(ab'c'd')(ab'cd)(abc'd)(a bcd') y1= (a'b'cd)(a'bc'd

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What must be subtracted from the polynomial x 3 1 3 x 2 3 5 x 2 5 so that the resulting polynomials is exactly divisible by x 2 1 1 x 1 0 View solution DivideShare It On Facebook Twitter Email 1 Answer 1 vote answered Dec 30, 18 by aditi (757k points) selected Dec 31Nên (a b c) 2 < a 2 b 2 c 2 (a 2 b 2) (a 2 c 2) (b 2 c 2) = 3(a 2 b 2 c 2) bởi Ngọc Hân 07/11/18 Like (0) Báo cáo sai phạm Cách tích điểm HP Nếu bạn hỏi, bạn chỉ thu về một câu trả lời

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(C) A family or household member who is an endangered adult (as defined in IC ) (d) The offense described in subsection (a)(1) or (a)(2) is a Level 4 felony if it results in serious bodily injury to a family or household member who is an endangered adult (as defined in IC )2a2( bc) (c) (ab2) 3c2 = e 3(a2b)2c = 0 14For group elements a, b, and c, express (ab)3 and (ab 2c) 2 without parentheses (ab)3 = ababab (ab 2c) 2 = ((ab 2c) 1)2 = (((ab 2)c) 1)2 = (c 1(ab 2) 1)2 = (c 1(b 2) 1a 1)2 = (c 1b2a 1)2 = c 1b2a 1c 1b2a 1 15Let G be a group and let H = fx 1 jx 2Gg Show that G = H as sets Here is a standardWhat must be subtracted from the polynomial x 3 1 3 x 2 3 5 x 2 5 so that the resulting polynomials is exactly divisible by x 2 1 1 x 1 0 View solution Divide

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Definition The longest side of the triangle is called the "hypotenuse", so the formal definition isGroup 1 (ac) • (a 2 c 2) Group 2 (ab) • (a 2 b 2) Group 3 (bc) • (b 2 c 2) Looking for common subexpressions Group 1 (ac) • (a 2 c 2) Group 3 (bc) • (b 2 c 2) Group 2 (ab) • (a 2 b 2) Bad news !!A and b are the other two sides ;

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There are various student are search formula of (ab)^3 and a^3b^3 Now I am going to explain everything below You can check and revert back if you like you can also check cube formula in algebra formula sheet a2 – b2 = (a – b)(a b) (ab)2 = a2 2ab b2 a2 b2 = (a –If a,b,c are all nonzero and a b c = 0, prove that a2/bc b2/ca c2/ab = 3 asked Sep 14, 18 in Class IX Maths by aditya23 ( 2,139 points) polynomials(3) A diagnostic device, if the sponsor complies with applicable requirements in § (c) and if the testing (i) Is noninvasive, (ii) Does not require an invasive sampling procedure that presents significant risk, (iii) Does not by design or intention introduce energy into a subject, and

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Grade 2 Module 3 Place Value, Counting, and Comparison of Numbers to 1,000 In this 25day Grade 2 module, students expand their skill with and understanding of units by bundling ones, tens, and hundreds up to a thousand with straws Unlike the length of 10 centimeters in Module 2, these bundles are discrete sets One unit can be grabbed and既に「a^2b^2c^2=3 から a^3b^3c^3≦3 は証明できない」と指摘したはずなんだけど 過去の投稿を探しましたが、見つけることができなく失礼しました。 次のようにも考えてみましたが、どうでしょうか。 abc=p,abbcca=q,abc=r とおく。a^2b^2c^2=3 は p^22q=3 となる。

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