[最も選択された] (x-y)^3 (y-z)^3 (z-x)^3 factorise 499744-(x-y)^3+(y-z)^3+(z-x)^3 factorise

Factorise (xy)^3 (yz)^3 (zx)^3 1976 3 View Full Answer Sriman Mishra, added an answer, on /9/15 Sriman Mishra answered this x 3 (yz) 3 y 3 (zx) 3 z 3 (xy) 3 = (xyxz) 3 (yzxy) 3 (zxzy) 3 = 3xyyzxz = 3 (xyz) 2 (By using a 3 b 3 c 3 3abc identity) Was this answer helpful?Solve x y z = x 3 y 3 z 3 = 8 in Z First I tried to transform this equation, substituting x = 8 − y − z So I end up with Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got z ∈ Z, which implies z − 8 is an integer implying that the second term is also an integer

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(x-y)^3+(y-z)^3+(z-x)^3 factorise

(x-y)^3+(y-z)^3+(z-x)^3 factorise-Click here👆to get an answer to your question ️ Factorise (x y)^3 (y z)^3 (z x)^3Solve 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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Each of the constituent expressions is separate, and it can be viewed as substituent math2x y z a/math math2y z x b/math math2z x y c/math Knowing this, you can use the formula for the sum of cubes matha^3 b^3 cSimilarly, ( y − z) and (zx) are factors So, (xy) (yz) (zx) is a factor In given expression, if we try to find terms with powers of x, x 3 term gets cancelled For x 2 we get − 3 x 2 y and 3x^2z Hence, coefficient for x 2 is 3 (yz) For the identidied factor we get term with x 2 as − x 2 ( y − z)Click here👆to get an answer to your question ️ Factorize x^3 y^3 z^3 = 3xyz

If xyz=6 and xyyzzx=12 then show that x 3 If xyz=6 and xyyzzx=12 then show that x 3 y 3 z 3=3xyz pls send the answer fast Please scroll down toFactor x^3(yz)y^3(zx)z^3(xy) Mathematics Posted by admin QUESTION POSTED AT 0936 PM Answered by admin AT 0936 PM This problem is unanswerable and i do know what i am talking about because i am getting my masters degree and i am not dumb Post your answerGet the answer to this question and access a vast question bank that is tailored for students

Equations Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations x^3(yz)^3y^3(zx)^3z^3(xy)^3 soClick here👆to get an answer to your question ️ Factorise 27x^3 y^3 z^3 9xyz Join / Login maths Factorise 2 7 x 3 y 3 z 3 Correct option is A (3 x y z) (9 x 2 y 2 z 2 − 3 x y − y z − 3 z x) We have, 2 7 x 3 y 3 z 3 if x1/x=5,then find value of x^31/x^3 The valuesof 249square 248square is 729X3512y3 Factorise (abc)³a³b³c3 I need very urgently please answer as quickly as you can Experts, please help me with the following questions attached below in the image Questions are from chapter POLYNOMIALS, grade 9 (please answer all of them

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We know the corollary if abc = 0 then a3 b3 c3 = 3abcUsing the above identity taking a = x−y, b = y−z and c= z−x, we have abc= x−yy−zz −x= 0 then the equation (x− y)3 (y−z)3 (z−x)3 can be factorised as follows(x−y)3 (y−z)3 (z−x)3 = 3(x−y)(y−z)(z−x)Hence, (x−y)3 (y−z)3 (z −x)3 = 3Correct answer Factorise x^3(yz)^3y^3(zx)^3z^3(xy)^3 answer fast 1mile equals approximately 16 kilometers which best approximates the number of miles in 6 kilometers?Algebra Factor (xy)^3 (xy)^3 (x y)3 − (x − y)3 ( x y) 3 ( x y) 3 Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 abb2) a 3 b 3 = ( a b) ( a 2 a b b 2) where a = xy a = x y and b = x− y b = x y

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(xyz)^3 put xy = a (az)^3= a^3 z^3 3az ( az) = (xy)^3 z^3 3 a^2 z 3a z^2 = x^3y^3 z^3 3 x^2 y 3 x y^2 3(xy)^2 z 3(xy) z^2 =x^3 y^3 z^3 3 xM a student solved this problem and said the answer is 3 feetMath, 0500, Simrankanojia Factorise (2xyz)^3 = (2yzx)^3=(2zxy)^3

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To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Factorize (i)`(ab)^3(bc)^3(ca)^3 `(ii) `x^3(yz)^3y^3(zx)^3z^3(xy)^3`Algebra Factor x^3y^3z^3 x3y3 z3 x 3 y 3 z 3 Rewrite x3y3 x 3 y 3 as (xy)3 ( x y) 3 (xy)3 z3 ( x y) 3 z 3 Since both terms are perfect cubes, factor using the sum of cubes formula, a3 b3 = (ab)(a2 −abb2) a 3 b 3 = ( a b) ( a 2 a b b 2) where a = xy a = x y and b = z b = z (xyz)((xy)2 −(xy)zz2) ( x y z) ( ( x yTrying to factor by pulling out 52 Factoring x 3 y x 3 z xy 3 xz 3 y 3 z yz 3 Thoughtfully split the expression at hand into groups, each group having two terms

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I'm trying to factorise $$ x^3z x^3y y^3z yz^3 xy^3 xz^3 $$ into four linear factors By plugging it into WolframAlpha I've learned that it's $$(xy)(xz)(yz)(xyz)$$ My question is what are the steps involved in factorising the expression? Supritha, added an answer, on 3/10/15 Supritha answered this It is form of (a)^ 3 (b)^ 3 (c)^3 =3abc where abc=0 a=xy , b=yz , c=zx abc= 0 = (xy) (yz) (zx)=0 = xyyzzx=0 = 0 =0 (a)^ 3 (b)^ 3 (c)^=3abc We know that x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz zx) Putting x y z = 0, x3 y3 z3 3xyz = (0) (x2 y2 z2 xy yz zx) x3 y3 z3 3xyz = 0 x3 y3 z3 = 3xyz Hence proved Show More Ex 25 Ex 25, 1 Ex 25,2 Important Ex 25,3 Important Ex 25,4

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Incoming Term: (x-y)^3+(y-z)^3+(z-x)^3 factorise, without finding the cubes factorise (x-y)^3+(y-z)^3+(z-x)^3, factorise the polynomial (x – y)3 + (y – z)3 + (z – x)3, factorise (2x-y-z)^3+(2y-z-x)^3+(2z-x-y)^3, 11.factorise (2x-y-z)^(3)+(2y-z-x)^(3)+(2z-x-y)^(3), factorise (x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3,