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Cho a,b,c là ba cạnh của tam giácC/m$(abc)(bca)(cab)\leq abc$ posted in Bất đẳng thức và cực trị Cho a, b, c là độ dài ba cạnh của tam giác Chứng minh a) $(abc)(bca)(cab)\leq abc$ b) $\frac{a}{bca}\frac{b}{acb}\frac{c}{abc}\geq 3$Nếu bạn hỏi, bạn chỉ thu về một câu trả lời Nhưng khi bạn suy nghĩ trả lời, bạn sẽ thu về gấp bội!Example Solve 8a 3 27b 3 125c 3 30abc Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 30abc) can be written as (2a) 3 (3b) 3 (5c) 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 3abc

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(a+b-2c)^3+(b+c-2a)^3+(c+a-2b)^3-Learn with Tiger how to do (a/b)^3(b/c)^3(c/a)^33 fractions in a clear and easy way Equivalent Fractions,Least Common Denominator, Reducing (Simplifying) Fractions Tiger Algebra Solver3(bc)(ca)(ab) If we use the following Result, we can immediately factorise the given Exp=3(bc)(ca)(ab) Result xyz=0 rArr x^3y^3z^3=3xyz Otherwise, consider the following Let, u=bc, v=ca rArr uv=ba=(ab) Now, The Exp=u^3v^3{(uv)}^3, =u^3v^3(uv)^3, =u^3v^3{u^3v^33uv(uv)}, =3uv(uv)=3uv{(uv)}, "

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X^3y^3z^3 = (xyz)(x^2y^2z^2xy xz yz) 3xyz is a identity/ You may prove by simple foiling for x = a^(1/3) and y = b^(1/3) and z = c^(1/3) you getA3 b3 c3 = (a b c) (a2 b2 c2 – ab – bc – ca) 3abcIf (ab)(bc)(ca)= 233 and (abc)= 16, then the value of c is CORRECT ANSWER 8 Enter the code shown above (Note If you cannot read the numbers in the above image, reload the page to generate a new one)

Question to simplify(a 2b 2) 3 (b 2c 2) 3 (c 2a 2) 3 / (ab) 3 (bc) 3 (ca) 3 Answer Let's solve the numerator first if abc = 0 Then a 3 b 3 c 3 = 3abc Now,(a 2b 2) 3 (b 2c 2) 3 (c 2a 2) 3 = a 2b 2 b 2c 2 c 2a 2 = 0 That means we can use this formula here because abc is coming out to be zeroHowever, there is a problem when trying to prove (abc)^2 ≤ 3(abbcca), because, in fact, the opposite is true (abc)^2 ≥ 3(abbcca) You can see that if you expand (abc)^2, simplify, multiply by 2, and use the trivial inequalityX^3y^3z^3 = (xyz)(x^2y^2z^2xy xz yz) 3xyz is a identity/ You may prove by simple foiling for x = a^(1/3) and y = b^(1/3) and z = c^(1/3) you get

3a^3 = 3a^3, a true equation Thus, a = b = c is always a solution, and abc = 3a = 3b = 3c Also, c = a b is a solution a^3 b^3 (a b)^3 = a^3 b^3 a^3 3a^2*b 3a*b^2 b^3 =3a^2*b 3a*b^2 =a*(3ab) b*(3ab) = 3ab * (a b) = 3abc Then abc = ab(ab) = 0Then take c=a and also sheres on ca(abc)^3a^3b^3c^3 =k(ab)(bc)(ca) K is unknown, so find her If a=1,b=1,c=0》(110)^31^31^30=k (11)(10)(01) =k*2 K=3 (abc)^3a^3b^3c^3 =3 (ab)(bc)(ca) October 29, 15 at 841 PM Unknown said See this one((ab)c)^3a^3b^3c^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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This fulllength music video of ABCmouse's cover version of "ABC" by The Jackson 5 features moreA^3(bc)b^3(ca)c^3(ab) This deals with simplification or other simple resultsAll equal mod 3, in which case the second factor is clearly divisible by 3, or A;B;C are all di erent mod 3, in which case (A 2B) (B C) 2 (C A) is equal to 1 1 1 modulo 3 and the second factor is again divisible by 3

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Click here👆to get an answer to your question ️ Prove that (a b c)^3 a^3 b^3 c^3 = 3(a b ) (b c) (c a)A/3 = 3 (B C)/3 = (BC) = BC Now, from here it's pretty easy You want to get B by itself so you want to get C on the other side of the equation Remember whatever you do on one side, you must do on the other side So, subtract C from both sides (A/3) C = B C C = B There is your answer B = (A/3) CThis fulllength music video of ABCmouse's cover version of "ABC" by The Jackson 5 features more

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For example, we can write $(ABC)' \equiv \big(A (BC)\big)' \equiv \big(A' \cdot (BC)'\big) \equiv A'\cdot (B'C') \equiv A'B'C'$ Indeed, provided we have a negated series of multiple variables all connected by the SAME connective (all and'ed or all or'ed), we can generalize DeMorgan's to even more than three variables, again, due to theChứng minh rằng (abc)^3 = a^3 b^3 c^3 3(a b)(b c)(c a) với mọi a, b, c bằng 3 cách Cách số 1 Khai triển vế trái thành vế phảiExample 1 Solve (4p 5q 3r) 2 Solution This proceeds as Given polynomial (4p 5q 3r) 2 represents identity first ie (a b c) 2 Where a = 4p, b = 5q and c = 3r Now apply values of a, b and c on the identity ie (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca and we get (4p 5q 3r) 2 = (4p) 2 (5q) 2 (3r) 2 2(4p)(5q) 2(5q)(3r) 2(3r)(4p) Expand the exponential forms

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Learning can truly be as easy as ABC, 123, and DoReMi!Bài 3 Ta có \(a^2b^2c^2=3\ge abbcca\) ( tự cm bđt nha ) Áp dụng bất đẳng thức Schwarz ta có \(\dfrac{a^3}{bc}\dfrac{b^3}{ca}\dfrac{c^3}{abThe 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

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Prove (a b)^3 (b c)^3 (c a)^3 = 3(ab)(bc)(ca) without expanding using the Multivariable Factor Theorem Guest Aug 21, 18 0 users composing answers(ab) 3 (bc) 3 (ca) 3 Let, a = ab b = bc c = ca then, abc = abbcca = 0 therefore, a 3 b 3 c 3 =3abc =3(ab)(bc)(ca)If a = b = c, then we get a^3 a^3 a^3 = 3*a*a*a;

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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)Prove (a b)^3 (b c)^3 (c a)^3 = 3(ab)(bc)(ca) without expanding using the Multivariable Factor Theorem Guest Aug 21, 18 0 users composing answersSolve for c A=(abc)/3 Rewrite the equation as Multiply both sides of the equation by Cancel the common factor of Tap for more steps Cancel the common factor Rewrite the expression Move all terms not containing to the right side of the equation Tap for more steps

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Learning can truly be as easy as ABC, 123, and DoReMi!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)Your approach is intuitive and that was also the first thing I thought;

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Equations Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations (ab)^3(bc)^3(ca)^3(ab)(bc)(ca) so that you understand betterChapter 3 CAB ECC and AHA 10 updates changed the CPR sequence from ABC to CAB Often in the ABC method chest compressions were delayed With the new Compressions – Airway – Breathing method a victim receives compressions faster, providing quicker critical blood flow to the vital organsAs 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 careers

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We can write \((abc)^3 = (abc)(abc)(abc) \) \(=>(abc)^3 = (abc)^2 (abc) \) we know that what is the formula of \( (abc)^2 \) \(=>(abc)^3 = (a^2b^2c^2 2ab 2bc 2ca) (abc) \)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 32Tiger was unable to solve based on your input (ab)3 (bc)3 (ca)3 Step by step solution Step 1 11 Evaluate (ca)3 = c33ac23a2ca3 Step 2 Pulling out like terms 21 (ab)^3 (ab)^32b^3 (ab)3 −(a −b)3 −2b3 https//wwwtigeralgebracom/drill/ (a_b)~3 (ab)~32b~3/

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Cách 1 Bạn nhóm $(ab)c^3$ lại rồi dùng hẳng đẳng thức phân tích như bình thường thôi P/s lát nữa mình edit bài viết ghi chi tiết sauThere 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 –A^3(bc)b^3(ca)c^3(ab) This deals with simplification or other simple results

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Please refer to the Explanation It is known that, (ab)^3=a^3b^33ab(ab) a^3b^3=(ab)^33ab(ab)(star) Setting, (ab)=d," we have, "a^3b^3=d^33abdI can do this better)if you take a=b,and you will see (bbc)^3(b)^3b^3c^3=c^3c^3=0,so (abc)^3a^3b^3c^3 shares on (ab) Bezout 's theoremthen take b=c and also sheres on bc Then take c=a and also sheres on ca(abc)^3a^3b^3c^3 =k(ab)(bc)(ca)Nếu bạn hỏi, bạn chỉ thu về một câu trả lời Nhưng khi bạn suy nghĩ trả lời, bạn sẽ thu về gấp bội!

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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 = 3abc Some not so Common FormulasThere 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 –Example Solve 8a 3 27b 3 125c 3 30abc Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 30abc) can be written as (2a) 3 (3b) 3 (5c) 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 3abc

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If a b c = 0, the prove that a × b = b × c = c × a where a, b, c are nonzero vectors View solution If a and b are vectors such that ∣ a b ∣ = 2 9 and a × ( 2 i ^ 3 j ^ 4 k ^ ) = ( 2 i ^ 3 j ^ 4 k ^ ) × b , then a possible value of ( a b ) ⋅ ( − 7 i ^ 2 j ^ 3 k ^ ) isClick here👆to get an answer to your question ️ In a Δ ABC , prove that sin^3 A cos(B C) sin^3 B cos(C A) sin^3 C cos(A B) = 3sin Asin Bsin CSolution 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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Incoming Term: (a-b)^3+(b-c)^3+(c-a)^3, (a-b)^3+(b-c)^3+(c-a)^3 is equal to, (a-b)^3+(b-c)^3+(c-a)^3 solution, (a-b)^3+(b-c)^3+(c-a)^3 formula, the value of (a-b)3+(b-c)3+(c-a)3 is, (a+b-2c)^3+(b+c-2a)^3+(c+a-2b)^3, (a+b-c)^3+(a-b+c)^3-8a^3, a+b+c=1 a^3+b^3+c^3=1, a+b=c a^3+b^3+3abc=c^3, a+b+c=0 then a^3+b^3+c^3=,