70以上 a^3+b^3+c^3-3abc formula proof 261432-A^3+b^3+c^3-3abc formula proof

Show activity on this post We have a3 b3 c3 − 3abc = det (a b c c a b b c a) = (a b c)(a ωb ω2c)(a ω2b ωc) where ω = exp2πi 3 is a primitive third root of unity On the other hand a b c = ex, a ωb ω2c = eωx, a ω2b ωc = eω2x and 1 ω ω2 = 0, hence a3 b3 c3 − 3abc = e0x = 1Adding 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) 2What is formula of a3b3 Math Algebraic Expressions and Identities NCERT Solutions;

A 3 B 3 C 3 3abc Formula

A^3+b^3+c^3-3abc formula proof

A^3+b^3+c^3-3abc formula proof-The 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 2So, a3 b3 c3 3abc = 0 (a b c ) (a2 b2 c2 ab ac bc) = 0 this means either (a2 b2 c2 ab ac bc) = 0 or (a b c ) = 0 (a2 b2 c2 ab ac bc) = 0 cannot be zero because 2a² 2b² 2c² 2ab 2ac 2bc = 0 a² b² 2ab a² b² 2c² 2ac 2bc = 0

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Fermat's Last Theorem is a theorem in number theory, originally stated by Pierre de Fermat in 1637 and proved by Andrew Wiles in 1995 The statement of the theorem involves an integer exponent n larger than 2 In the centuries following the initial statement of the result and before its general proof, various proofs were devised for particular values of the exponent n= a 3 b 3 3ab(a b) On solving it further we get a 3 b 3 3a 2 b 3ab 2 Hence, in this way we obtain the identity ie (a b) 3 = a 3 b 3 3ab(a b) = a 3 b 3 3a 2 b 3ab 2 Example 1 Solve (3a 2b) 3 Solution This proceeds as Given polynomial (3a 2b) 3 represents the identity (a b) 3 Where a = 3a and b = 2b NowThe left hand side proof is tricky but here it is, although it would be much easier to use the right hand side given a^3 b^3 c^3 3abc factor a^3 b^3 using cubic formula (ab) (a^2 ab

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 get = (2a 3b 5c) { (2a) 2 (3b) 2 (5c) 2 (2a)(3b) (3b)(5c) (5c)(2a) } Expand the exponential forms and we getA 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)Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the

Since the LHS is symmetric in a, b, c, and the RHS can be written as ( ( a b c) − 3 a 2) 3 it suffices to prove the inequality with the additional assumption a ≤ b ≤ c Replacing a, b, c by a, a x, a x y, respectively (where x, y ≥ 0 ), we getSo basically what it is is that the problem is a^3b^3=c^3 but you changed it to cube root a^3 cube rootb^3 =cube rootc^3 which is equal to ab=c, so if what you say is right you would be able to use any terms for this second equation and it would fit the third so lets say 11=2 then plug in you get 11=8 doesn't work753 x 753 x 753 247 x 247 x 247

If A 1 3 B 1 3 C 1 3 0 Then A A B C 0 B A B C 3 27 A B C C A B C 3 Youtube

What Are Various Forms To Write A B C A Whole Cube Quora

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 = 3abc2 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 32A^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

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Example 22 If a b c = 0, then prove that a 3 b 3 c 3 = 3abc Solution We know that Example 23 Find the following product (x y 2z) (x 2 y 2 4z 2 – xy – 2yz – 2zx) Solution We have, Example 24 If a b c = 6 and ab bc ca = 11, find the value of a 3 b 3 c 3 – 3abc SolutionIf A is an elementary matrix and B is an arbitrary matrix of the same size then det(AB)=det(A)det(B) Indeed, consider three cases Case 1 A is obtained from I by adding a row multiplied by a number to another row In this case by the first theorem about elementary matrices the matrix AB is obtained from B by adding one row multiplied by a number to another rowFurther more, note that the yellow part is a cuboid with height a, width a, and length b ;

What Is The Formula Of Math A 3 B 3 Math Quora

What Is The Formula Of Math A 3 B 3 Math Quora

Thx Socratic Prove that a3 b3 c3 − 3abc = (a b c)(a2 b2 c2 − ab − bc − ca)This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product As a special case, the square of a triple product is a Gram determinant;A 3 ± b 3 = (a Same sign b)(a 2 Opposite sign ab Always Positive b 2) Note The quadratic portion of each cube formula does not factor, so don't waste time attempting to factor it Yes, a 2 – 2ab b 2 and a 2 2ab b 2 factor, but that's because of the 2 's on their middle terms

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Ashutosh Verma, Meritnation Expert added an answer, on 24/7/14 Ashutosh Verma answered this Answer Factorise a 3 b 3 c 3 3abc We know ( a 3 b 3 ) = ( a b ) ( a 2 ab b 2 ) So we rearrange it As ⇒ ( a b ) ( a 2 ab b 2 ) c 3 3abc ⇒ ( a b ) ( a 2 2 ab b 2 3ab) c 3 3abcUsing the formula, a^3 b^3 c^3 = (a b c) (a^2 b^2 c^2 ab bc ca) 3abcIf abc=0,then a^3 b^3 c^3 = 3abcabca3b3c3 = 3abca3 abcb3 abcc3 = 3bca2 cab2 abc2 = 3Answer (C) 3 Answer verified by Toppr Upvote (70)Thx Socratic Prove that a3 b3 c3 − 3abc = (a b c)(a2 b2 c2 − ab − bc − ca)

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