X^2 y^2 z^2-xy-yz-zx formula 237285

Apply the formula for (ab)²=a²b²2ab for (xy)²,(yz)² & (zx)² and add them together toThe electric potential existing in space is `V(x,y,z)A(xyyzzx)` (a) Write the dimensional formula of A (b) Find the expression for the electric fieAnswered 4 years ago x^2y^2z^2xyyzzx= { (xy)^2} { (yz)^2} { (zx)^2}÷2 sum of squares of three real nos is zero if each is equal to zero so, xy=0, yz=0, zx=0 therefore, x=y=z 23K views Supriyo Ain , Studying in Class 11 Mathematics & Relativity, Hindmotor High School

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X^2 y^2 z^2-xy-yz-zx formula

X^2 y^2 z^2-xy-yz-zx formula-Factor x^2y^2 x2 − y2 x 2 y 2 Since both terms are perfect squares, factor using the difference of squares formula, a2 −b2 = (ab)(a−b) a 2 b 2 = ( a b) ( aRf(P) = h6;10;8i The tangent plane at P has

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Let us consider LHS of the equation LHS = x 3 y 3 z 3 – 3xyz LHS = 1 3 2 3 3 3 – 3(1 × 2 × 3) LHS = 1 8Use the formula a 3 b 3 c 3 − 3 a b c = ( a b c) ( a 2 b 2 c 2 − a b − b c − c a) = 1 2 ( a b c) ( ( a − b) 2 ( b − c) 2 ( c − a) 2) and then notice that a − b = x 2 − y 2 z x − y z = ( x − y) ( x y z) This will lead you to get the answer as ( x 3 y 3 z 3 − 3 x y z) 2 as pointed out by Ewan Delanoy Share Follow this answer to receive notificationsJACOBIAN Find ꝺ (u,v,w)/ꝺ (x,y,z) where u=x^2 y^2 z^2 v=xyyzzx w=xyz Watch later Share Copy link Info Shopping Tap to unmute If playback doesn't begin shortly, try restarting

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, music∂z ∂x = x(x2 y3)9 1 x (Note We used the chain rule on the first term) ∂z ∂y = 30y 2(x y3)9 (Note Chain rule again, and second term has no y) 3 If z = f(x,y) = xexy, then the partial derivatives are ∂z ∂x = exy xyexy (Note Product rule (and chain rule in the second term) ∂z ∂y = x2exy (Note No product rule, but weSolutionShow Solution x y z = 4 and x 2 y 2 z 2 = 30 Since ( x y z)2 = x2 y2 z2 2 ( xy yz zx ), we have (4)2 = 30 2 ( xy yz zx ) ⇒ 16 = 30 2 ( xy yz zx ) ⇒ 2 ( xy yz zx ) = 14 ⇒ xy yz zx = `14/2` = 7 ∴ xy yz zx = 7 Concept Expansion of Formula

, z 1) and (x 2, y 2, z 2) parallel to the coordinate planes The length of edges are x 2 – x 1, y 2 – y 1, z 2 – z 1 and length of diagonal is 2 2 2 ( ) ( ) ( )x x y y z z2 1 2 1 2 1− − − 1215 Section formula The coordinates of the point R which divides the line segment joining two points P(x 1, y 1, z 1) and Q(x 2, y 2, zX³ y³ z³ –3xyz = (xyz)(x 2 y 2 z 2 –xy–yz−xz) x ² y ² =12(xy) ² (x–y) ² (xa)(xb)(xc)=x³ (abc)x² (abbcca)xabc x³ y³ = (xy)(x² – xy y²) x³ – y³=(x–y)(x² xy y²) x² y² z² − xy – yz – zx = 12(x−y)² (y−z)² (z−x)²Arrange the expression in the form of factorization (x y z)(xy yz zx)− xyz ( x y z) ( x y y z z x) − x y z Expand the expression x2y x2z xy2 2xyz xz2 y2z yz2 x 2 y x 2 z x y 2 2 x y z x z 2 y 2 z y z 2 Do factorization (x y)(x z)(y z) ( x y) ( x z) ( y z)

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Multiply X2 Y2 Z2 − Xy Xz Yz By X Y − ZFor example, x 2 zp y 2 zp = xy and (x 2 – yz) p (y 2 – zx) q = z 2 – xy are first order quasilinear partial differential equations Nonlinear equation A first order partial differential equation f(x, y, z, p, q) = 0 which does not come under the above three types, in known as a nonliner equationZ Z S F~ ·dS~ = R R D (−P ∂g ∂x − Q∂g ∂y R)dA = R1 0 R1 0 −(xy)(−2x)− (yz)(−2y)(zx)dxdy = R1 0 R1 0 −(xy)(−2x)− y(4−x2 − y2)(−2y)(4−x2 − y2)xdxdy R1 0 R1 0 4x− x3 2x2y 8y2 − xy2 − 2x2y2 −2y4dxdy = 713/180 Section177, #24Evaluate the surface integral

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Given x y z = 6, x2 y2 z2 = 10 x3 y3 z3 = 12 Formula used (x y z)2 = x2 y2 z2 2(xy yz zx) x Q1 Columbus started his journey from Lucknow to Kolkata which is 0 km, at the speed of 40 km/h then he went to Banglore which is 300 km at the speed of km/hThis gives an implicit formula of x 2 y 2 y 2 z 2 z 2 x 2 − r 2 x y z = 0 {\displaystyle x^ {2}y^ {2}y^ {2}z^ {2}z^ {2}x^ {2}r^ {2}xyz=0\,} Also, taking a parametrization of the sphere in terms of longitude (θ) and latitude (φ), gives parametric equations for the Roman surface as follows First We take RHS & use the Formula ( ab)²= a²b²2ab & simplify it then RHS becomes equal to LHS RHS ⇒ 1/2×(x y z) (x² y²2xy y² z²2yzx²z²2xz)

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Neitherwaseven x=2h1and y=2k1 Then z 2=4h 4h14k2 4k1=4(h hk2 k)2 Thissaysthat2z2 but4 ∤ z2 whichisimpossiblebecause z2 isaperfectsquare Weshallchoose xtobeeven Theorem The triple (x,y,z) is a primitive Pythagorean triple if and only if there existtworelativeprimeintegers sand tsothat s > t >0and x=2st y= s2 −t2 and z= s2 t2X^22*x*yy^2z^2x^2y^2z^2–2*x*y2*x*z2*y*z= 2*z^22*x*z2*y*z which is not identically 0, so there are a set of points where the 2 expressions are equal (such as z=0 and z=xy) but in general they are not equal views View upvotes 9 Related Answer 19 x 2 y 2 z 2 − xy – yz –zx = 1/2(x − y) 2 (y − z) 2 (z − x) 2 Maths Algebraic Identities For Class 9 Do you know the difference between an algebraic formula (identity) and an algebraic expression?

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An algebraic formula is an equation that is a rule written using mathematical and algebraic symbols (a b) 2 =a 2 Making m = xy n = yz p = xz we have m^2n^2p^2 = 2(x^2y^2z^2x*yx*zy*z) then x^2y^2z^2x*yx*zy*z = 1/2((xy)^2(yz)^2(xz)^2) Algebra ScienceX^2y^2z^2xyyzzx=0 multiplying the RHS and LHS by 2 we get , 2 x^2y^2z^2xyyzzx =0 or, (xy)^2(yz)^2(zx)^2=0 since in LHS there are only squared terms,ie they cannot be negative What are factors of 4x^212xy9y^24x6y35?

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M4maths Previous Puzzles Which one statement is correct If x2 y2 z2 xy yz zx 0 then which one statement is correct OPtion 1) x = y = z 2) x > y > z 3) x y z 4) x ≠ y = z 5) x = y ≠ z 6) x ≠ y ≠ zP z Q O x y C D E Q(x2;y2;z2)is dist(P;Q)= q (x1¡x2) 2(y 1¡y2) 2(z 1¡z2) 2 We shall use the dimension reduction to verify this formula In the above &gure, we project point P and Q vertically (parallel to z ¡axiz) onto xy ¡Click here👆to get an answer to your question ️ Using the identity and proof x^3 y^3 z^3 3xyz = (x y z)(x^2 y^2 z^2 xy yz zx)

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Answer (1 of 12) xyz=0, so xy= z; Example Ifu =x y z, v = x y2 z2 and w = yz zx xy prove that (grad u) {grad v x grad w} = 0 dear frist of all u solve grad U ,grad V ,grad W after thenu get further solutionwhich is already attached by others tutor then if x y z = 5 x 2 y 2 z 2 = 21 y 2 = zx Formula used (x y z) 2 = x 2 y 2 z 2 2(xy yz xz) Calculations (x y z) 2 = 5 2 ⇒ x 2 y 2 z 2 2(xy yz xz) = 25 ⇒ 21 2(xy yz xz) = 25 ⇒ 2(xy yz xz) = 25 21 ⇒ xy yz xz = 4/2 ⇒ xy yz xz = 2 ⇒ xy yz y 2 = 2 ⇒ y(x y z) = 2 ⇒ y(5) = 2 ⇒ y = 2/5

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Xz= y = (x^2/yz) (y^2/xz) (z^2/xy) = x^2(xz)(xy) y^2(yz)(xy) z^2 (yz)(xz) / (yz)(xz)(xy) = x^2 (x^2yz) y^2 (xy^2z) z^2 (xyz^2) / x^2y^2z^2 = x^4yz xy^4z xyz^4 / x^2y^2z^2 = xyz (x^3 y^3 z^3) / (xyz)(xyz) = x^3#2 in 116 Find the equation of a tangent plane and the equation of a normal line to the surface x2 y2 z2 = 18 at the point P(3;5; Important Statistics Formulas (I) The Mean of Grouped Data can be found by 3 methods Direct Method formula This method can be very calculation intensive if the values of f and x are largeWe have big calculations and chance of making mistake is quite high 2 Assumed mean method formula Where a= Assumed mean and d i = x i –a This method is quite useful

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How do I prove that x^2y^2z^2xyyzzx=1\2 ((xy) ^2(yz) ^2(zx) ^2)?We rst note that these planes intersect along the line x y= 1 It follows that the base of Eis a 2D region Dthat can be described by the inequalities x 0, y 0, and x y 1 Given x^2y^2z^2=xyyzzx (1)multiplying by (xyz) on both sidesx^2y^2z^2 (xyz)=(xyyzzx ) (xyz)expand the above equationsx^3y^3z^3xy^2x^2 yyz^2y^2 zxz^2x^2 z=x^2 yxyzzx^2xy^2 y^2 zxyzxyz

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Example 2 if x = 10 and y is 4 (10 4) 2 = 10 2 2·10·4 4 2 = 100 80 16 = 36 The opposite is also true 25 a 4a 2 = 5 2 2·2·5 (2a) 2 = (5 2a) 2 Consequences of the above formulas4) Let f = x2y2 z2 Then the surface is a level surface of f Therefore, the gradient of f at P is normal to the surface We compute this vector rf = h2x;2y;The formula of x 3 y 3 z 3 – 3xyz is written as \(x^{3} y^{3} z^{3} – 3xyz = (x y z) (x^{2} y^{2} z^{2} – xy – yz – zx)\) Let us prove the equation by putting the values of x = 1;

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(2) ∂f ∂z = lim ∆z→0 f(x,y,z ∆z)−f(x,y,z) ∆z (3) These formulae are direct generalisations of the well known definition of the derivative of a function f(x) of one variable x df dx = lim ∆x→0 f(x∆x)−f(x) ∆x (4) Example Let f(x,y,z) = x2yz yez, then ∂fF(x,y,z) = x2yz −xy2z decreases in the y direction (a) (1,−1,2), (b) (1,1,1), (c) (−1,1,2), (d) (1,0,1) Definition if nˆ is a unit vector, then nˆ·∇f is called the directional derivative of f in the direction nˆ The directional derivative is the rate of change of f in the direction nˆ11 Functions of Two or More Variables A symbol z which has a definite value for every pair of values of x and y is called a function of two independent variables x and y and is written as z = f (x, y) or I (x, y) 12 Limits "The function f (x, y) is said to tend to limit l as x oa and y o b if and only if the limit l is independent of the path followed by the point (x, y) as x o a and y ob

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X^2y^2z^2xyyzzx=0 multiplying the RHS and LHS by 2 we get , 2 x^2y^2z^2xyyzzx =0 or, (xy)^2(yz)^2(zx)^2=0 since in LHS there are Ex 42, 9 By using properties of determinants, show that 8 (x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z – x) (xy yz zx) Solving LHS 8 (𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Applying R1→ R1 – R2 = 8 (𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Ex 42, 9 By using properties of determinants, show that 8 (x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z –(B)Factorize the polynomial using the factorization formula  10 1 6x 9x2 11 144x2 – 72x 9 12 4a2b2 abcd 25c2d2 13 x2 y2 – a2 – b2 2xy 2ab

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Or x = c 1 y ie, c 1 = x / y From the last two ratios, Integrating, log y = log z log c 2 or y = c2 z ie, c2 = y / z Hence the required general solution is Φ( x/y,= 0,y/z)where Φ is arbitrary Example 22 Solve p tan x q tan y = tan z The subsidiary equations are x2y2z2xyyzzx=0 However, the graph is a line x=y=z as can be seen by using the CauchySchwarz inequality (a12 an2)(b12 bn2) ≥ (a1b1 anbn)2 With equality happening if for all 1 ≤ k ≤ n, ak= bk In this case, take (a1,a2,a3) = (x,y,z) and (b1,b2,b3) = (y,z,x) (x2 y2 z2)(y2 z2 x2) ≥ (xy yz zx)2X^2y^2z^2xyyzzx formula X^2y^2z^2xyyzzx formulaFactor x^2y^2 x2 − y2 x 2 y 2 Since both terms are perfect squares, factor using the difference of squares formula, a2 −b2 = (ab)(a−b) a 2 b 2 = (a b) (a b) where a = x a = x and b = y b = y (xy)(x−y) (x y) (x y)(x y) 2 = x 2 y 2 2ab Therefore, we can write the above equation as;

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 x2y2z2=xyyzzx you can get 2x22y22z2=2xy2yz2zx which equals 2x22y22z22xy2yz2zx=0,the equivalant is (xy)2(yz)2(zx)2=0If $xy yz zx = 1$, then show that $$\dfrac{x}{1x^2} \dfrac{y}{1y^2} \dfrac{z}{1z^2} = \dfrac{4xyz}{(1x^2)(1y^2)(1z^2)}$$ I tried doing the sum algebraically, that is, by solving Stack Exchange NetworkAlgebra Examples Rewrite (xy z)2 ( x y z) 2 as (xyz)(xyz) ( x y z) ( x y z) Expand (xyz)(xyz) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression Simplify each term Tap for more steps Multiply x x by x x Multiply y y by y y

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= y z x y ¯ z x y z ¯ = x ( y X O R z) y z, so the formula is symmetric in permutations of x,y,z, and the expression is true at least whenever two of the variables are true However, it's fairly easy to see that it is false when at most one variable is true

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