Factorize (x-y)^3 (y-z)^3 (z-x)^3 659321-Factorize (2x-y-z)^3+(2y-z-x)^3+(2z-x-y)^3

 If x y z = 0, show that x^3 y^3 z^3 = 3xyz (with Video) Ex 25, 13If x y z = 0, show that x3 y3 z3 = 3xyz 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 proAns 1, 3, 4 15 The relation R on A = {x,y,z} where R = {(x,x),(y,z),(z,y)} Ans 2 16 The relation R on Z where aRb means a ≠ b Ans 2 17 The relation R on Z where aRb means that the units digit of a is equal to the units digit of b Ans 1, 2, 4 18 The relation R on N where aRb means that a has the same number of digits as b Ans factorize x y 3 y z 3 z x 3 Mathematics TopperLearningcom 3z3pm4pp Practice Test MCQs test series for Term 2 Exams ENROLL NOW

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Factorize (2x-y-z)^3+(2y-z-x)^3+(2z-x-y)^3

Factorize (2x-y-z)^3+(2y-z-x)^3+(2z-x-y)^3-Factor(x^3y^3z^3 ) Natural Language;Answer (1 of 2) Expression = (xy)(x y) (yz)(y z) (zx)(z x) 2(x)(y)(z) = (xy)(x y z) (yz)(x y z) (zx)(z x) = (xy yz)(x y z) (zx)(z x

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Capítulo 1 Factorización 11 Factorizar las siguientes expresiones algebraicas como producto de un binomios y un trinomio 1 4x14y xy y2Sol (8y)(6 xy) 2 −28−4x−4xy 7z xz xyzSol (7xxy)(−4z) 3 18−6x−3xy x2y −3yz xyzSol (−3x)(−6xy yz)4 10x2 x3y 10xz 10yz x2yz xy2zSol (10xy)(x2 xz yz) 5 18−9z −3yz z2 yz2 Use sum of cubes identity to find x^3y^3z^3 = (xyz)(x^2y^2xyzz^2) Use the sum of cubes identity a^3b^3=(ab)(a^2abb^2) with a=xy and b=z as follows x^3y^3z^3 =(xy)^3z^3 =((xy)z)((xy)^2(xy)zz^2) =(xyz)(x^2y^2xyzz^2) Without finding cubes, factorise 1(xy)3 (yz)3 (zx)3 2(x2y)3 (2y3z)3 (3zx)3 Get the answers you need, now!

Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, musicSOLUTION 1 Begin with x 3 y 3 = 4 Differentiate both sides of the equation, getting D ( x 3 y 3) = D ( 4 ) , D ( x 3) D ( y 3) = D ( 4 ) , (Remember to use the chain rule on D ( y 3) ) 3x 2 3y 2 y' = 0 , so that (Now solve for y' ) 3y 2 y' = 3x 2, and Click HERE to return to the list of problems SOLUTION 2 Begin with (xy) 2 = x y 1 Differentiate both sides So it factors as $(xyz)(Ax^2 Ay^2 Az^2 Bxy Byz Bxz)$ Then let $x = 0$ and we get $(yz)(\text{something}) = y^3 z^3$ This is a wellknown factorization, and the students can easily tell you what 'something' is Then we quickly get $A = 1$, $B = 1$

To factorise z 3 1 into a linear and quadratic factor, it is best to try the factor theorem first We set the equation to zero z 3 1=0We need to find a value of z such that the above is true By inspection we can see that if z=1 then the above is true because (1) 3 1=0 since (1) 3 =1Therefore, (z1) is a factor The other factor is quadraticHere, the exponent of the second term, ie, x–1 is –1, which is not a whole number So, this algebraic expression is not a polynomial Again, x 3 can be written as 1 2 x 3 Here the exponent of x is 1 2, which is 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

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You can use an identity like a^3b^3=(ab)(a^2abb^2) for 2 of the cubes, and hope that it might end up having something in common with the third cube Let's see (2xyz) ^3 (2yzx) ^3= (2xyz2yzx)P(x,y,z)=(xy2zLaintegraldesuperficie Es una elipse en el plano z = y 1 Su proyección sobre el plano XY es la curva ° de ecuación x2 (y ¡1)22 = 1 (una elipse, también) Sea S la superficiedelplano z = y 1 limitadaporC;sepuedeparametrizarcomo r(x;y) = (x;y;y 1);Using the above identity taking a=x−y, b=y−z and c=z−x, we have abc=x−y y−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)

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= (x y)(x – y)(x 4 x 2 y 2 y 4), by (1) (b) x 6 – y 6 = (x 3) 2 – (y 3) 2 = (x 3 y 3)(x 3 – y 3), by (1) = (x y)(x 2 – xy y 2)(x y)(x 2 xy y 2), by (2) and (3) Which of the above factorization is correct?Expand polynomial (x3)(x^35x2) GCD of x^42x^39x^246x16 with x^48x^325x^246x16 factorise the following a) (xy) 3 (yz) 3 (zx) 3 b) 25x 2 16y 2 4z 2 40xy16yz zx c) 1 64x 3 d) 2root2a 3 8b 3 27c 3 18root2 abc e) 842r

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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` Now we need to factor (x^4 2x^3(yz) 2x^2(yz)^2 x(yz)^3 yz(y^2z^2yz) 1 This has a factor x y and also x z so we have a factor (x^2(yz)x yz) so let other factor be (x^2axc) (x^2(yz)x yz) (x^2axc) = x^4 x^3(ayz) x^2(ca(yz) yz) x(ayzc(yz)) cyz) 2 Comparing coefficients from 1 and 2 From x^3 a= yz,Z = x2y3 2x4y ∂z/∂x =2xy3 2 ∂z/∂y =3x2y2 4 • REMEMBER When you are taking a partial derivative you treat the other variables However, we can treat dy/dx as a fraction and factor out the dx dy = f0(x)dx where dy and dx are called differentialsIfdy/dx can be interpreted as

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Example 1 Factorize x 2 y 2 y 2 Solution Completely factorize x 2 y 2 y 2, ie write the greatest common factor of x 2 y 2 and y 2 as the common factor y 2 is the highest common factor To find the other factor, divide each term by the HCF and add the quotients (x 2 y 2)/y 2 = x 2 and y 2 /y 2 = 1 Sum of the quotients is x 2 1, which is the other factorTherefore,If f(z) = u(x;y)iv(x;y), where z = xiy, explain why f(z) is analytic and write a formula forf0(z) in terms of the derivatives of uand v Since the derivatives ux, uy, vx and vy are everywhere de ned and continuous and since the CauchyRiemann equations hold everywhere, by our basic theorem on di erentiability, we know that f(z) is di erAlgebra Factor x^3y^3 x3 − y3 x 3 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 = x a = x and b = y b = y

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