Prove that the equation $x^3y^3z^33xyz=1$ defines a surface of revolution and find the analytical equation of its axis of revolution I think that I need to apply Euler's formula, so that I g Stack Exchange Network x³ y³ z³ 3xyz = (x y z) (x² y² z² – xy – yz– zx) In order to find the formula of x³ y³ z³, we need to send 3xyz to the right side of equal sign Thus the formula becomes, x³ y³ z³ = (x y z) (x² y² z² – xy – yz– zx) 3xyzIf x y z = 6, x2 y2 z2 = 16 and x3 y3 z3 = 196 Then find the value of 3xyz 1 160 2 140 3 130 4 1
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X^3+y^3+z^3-3xyz formula-Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchangeयदि x y z = 9, xy yz zx = 23, तब `x^(3) y^(3) z^(3) 3xyz=` A 100 B 81 C 108 D 123 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries
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verify that ` x^3 y^3 z^3 3xyz = 1/2( xyz) (xy)^2 (yz)^2 (za)^2 ` Books Physics NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless Chemistry NCERT P Bahadur IITJEE Previous Year Narendra Awasthi MS Chauhan Biology NCERT NCERT Exemplar NCERT Fingertips 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 MoreThe formula of x 3 y 3 z 3 – 3xyz is written as \(\begin{array}{l}x^{3} y^{3} z^{3} – 3xyz = (x y z) (x^{2} y^{2} z^{2} – xy – yz – zx)\end{array} \) Let
Click here👆to get an answer to your question ️ Factorise 27x^3 y^3 z^3 9xyzX 3 y 3 z 3 3xyz = (xy) 3 3x 2 y 3xy 2 z 3 3xyz = (xy) 3 z 3 3xy (xyz) = (xyz) (xy) 2 z (xy) z 2 3xy (xyz) = (xyz) (x 2 2xyy 2 xzyzz 2 3xy) = (xyz) (x 2 y 2 z 2 xyyzzx) bởi Nguyễn Thanh Huyền Like (1) Báo cáo sai phạm Cách tích điểm HP What must be subtracted from 4x^42x^36x^22x6 so that the result is exactly divisible by 2x^2x1?
Answer is (xy z)(x^2 y^2 xyz z^2) You can check by multiplying it out Notice that each term is a perfect cube x^3 y^3 = (xy)^3 So we have a sum of cubes, and the factoring formula is a^3 b^3 = (ab)(a^2abb^2) So we use a = xy and b = z to get x^3 y^3 z^3 = (xy)^3 z^3 = ((xy) z)((xy)^2(xy)zz^2) =(xy z)(x^2 y^2 xyz z^2) check by multiplying itGiven x y z = 8, xy yz zx = 15 Formula used x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz Q1 There were 1 students who went for summer camp students joind them after some time, Due to which the cost of consumption increased by Rs 80 per day and the average cost of each students decrease by RsSo,x3y3z3−3xyz =(xyz)(x2y2z2−xy−yz−zx) = (xyz)((xyz)2−3xy−3yz−3zx) = (xyz)((xyz)2−3(xyyzzx) = 9(92−3×23) = 9(81−69) = 9×12 = 108 (a) 108 Mathematics Secondary School Mathematics IX Standard IX Suggest Corrections 0
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Given x y z = 0 Formula used x3 y3 z3 = {(x y z) × (x2 y2 z2 xy yz zx)} 3xyz Calculation &rA With hundreds of Questions based on Algebra, we help you gain expertise on Quantitative AptitudeAll for free Explore Testbook Learn to attain the subject expertise with usX^3y^3z^33xyz=(xyz)(x^2y^2z^2xyyzzx)a^3b^3c^33abc=(abc)(a^2b^2c^2abbcca)a^3b^3c^33abc formula proofx^3y^3z^33xyz formula proofaA) 36 b) 40 c) 42 d) 48
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Complete stepbystep answer Now, we will use partial differentiation It is denoted by ∂ We are given u = log (x3 y3 z3 3xyz) So, partially differentiating u with respect to x, we get ∂u ∂x = 3x2 3yz (x3 y3 z3 3xyz) Similarly, partially differentiating u with respect to y and z, we getAnswer by lenny460 (1073) ( Show Source ) You can put this solution on YOUR website! If ${ax by cz =1}$, then show that in general ${x^3 y^3 z^3 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3b^3c^33abc)}$, of
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Given (xyz) = 9∴ (x y z)2 = 92 = 81∴ x2 y2 z2 2 × 26 = 81∴ x2 y2 z2 = 81 52 = 29Now, x3 y3 z3 3xyz= (x y z) (x2 y2 z2) (xyFind the value of x3 y3 z3 3xyz if x y z 12 and x2y2z270 Hint Here, we have to find the value of the algebraic expression We will find the value of the sum of the product of the variables by substituting the given equation in the square of the sum of the three variables' identityThere are two formula of it 1 x^3 y^3 z^3 3xyz = (xyz) (x^2y^2z^2xyyzzx) 2 x^3 y^3 z^3 3xyz = (1/2) (xyz) {xy)^2(yz)^2(zx)^2}
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There are two formula of it 1 x^3 y^3 z^3 3xyz = (xyz) (x^2y^2z^2 xyyzzx) (xyz)(x^2y^2z^2xyyzxz) Stepbystep explanation and plz follow meConsider (x – y – z) 2 ⇒ x 2 (y 2) (z 2) 2 x (y) 2 (y) (z) 2 (z) x ⇒ (x – y – z) 2 = x 2 y 2 z 2 – 2xy – 2xz 2yz Let us prove the above equation Consider x = 2, y = 3 and z = 4 Substitute in the above equation we get LHS = (x – y – z)X y z = 2 x 3 y 3 z 3 3xyz = 74 Formula used a 3 b 3 c 3 3abc = (a b c)/2 3(a 2 b 2 c 2) (a b c) 2 Calculation According to the formula 2/2 3(x 2 y 2 z 2) 2 2 = 74 ⇒ 3(x 2 y 2 z 2) = 74 4 ⇒ x 2 y 2 z 2 = 78/3 = 26 ∴ Required answer is 26
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If the polynomial k 2 x 3 − kx 2 3kx k is exactly divisible by (x3) then the positive value of k is ____If x y z = 5 and xy yz zx = 1 then Find the value of (x 3 y 3 z 3 – 3xyz) (x y z) 2 1 125 2 135 3 133 4 130X^3 (yz)^3 y^3 (zx)^3 z^3 (xy)^3
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Formula x 3 y 3 z 3 3xyz = 1/2 (x y z) (x y) 2 (y z) 2 (z x) 2 Calculation According to the question ⇒ x 3 y 3 z 3 3xyz = 1/2 × (255 256 257) × (255 256) 2 (256 257) 2 (257 255) 2 ⇒ x 3 y 3 z 3 3xyz = (1/2) × 768 × (1) 2 (1) 2 (2) 2 ⇒ x 3 y 3 z 3 3xyz = (1/2) × 768 × 1 1 4If x y z = 6 and xy yz zx = 10, then the value of x3 y3 z3 3xyz is?(xyz)^3 (x y z) (x y z) (x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy x * z = xz y * x = xy
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Answer (1 of 5) I thought I give it a try as well I have added a mirror, so the depth is easier to comprehend It is a somewhat rounded version of But although the graph seems to be a curved table cloth, there are actually three straight lines embedded these three lines are of course (z(xyz) (x ^ 2 xy y ^ 2 xzyz z ^ 2) หลักฐาน โปรดทราบว่า x = y z เป็นคำตอบของ x ^ 3y ^ 3z ^ 33xyz = 0 เสียบ x = y z ในสมการข้างต้น (y z) ^ 3y ^ 3z ^ 33 (y z) yz = y ^ 3 3y ^ 2z 3yz ^ 2 z ^ 3 y ^ 3z ^ 33y ^ 2z3yz ^ 2 = 0 เราจึงสามารถหารSolution x = 2x, y = 2y and z = 4z If x y z = 0, then x 3 y 3 z 3 = 3xyz 8x 3 27y 3 64z 3 = 3 (2x) (2y) (4z) = 48xyz After having gone through the stuff given above, we hope that the students would have understood, "x cube plus y cube plus z cube minus 3xyz" Apart from the stuff given in this section, if you need any other
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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 x^2y 3xy^2 3 ( x^2 y^2 2xy ) z 3x z^2 3yz^2 =x^3y^3z^3 3x^2 y3xy^2 3x^2 z 3y^2 z 6xyz 3xz^2 3 yz^2 arrange in orderThen we get x 3 ( x y) 3 = ( x y z) 3 − 3 x ( y z) ( x y z) By using equation (1) we can simplify the RHS of the above equation And by using equation (2), we can further expand LHS of the above equation Hence, by using equation (1) and (2), weGiven 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 There were 1 students who went for summer camp students joind them after some time, Due to which the cost of consumption increased by Rs 80 per day and the average cost of each students decrease by Rs
X3 Y3 Z3 3xyz
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Show activity on this post The answer is yes, the rational points on your surface lie dense in the real topology Let's consider the projective surface S over Q given by X3 Y3 Z3 − 3XYZ − W3 = 0 It contains your surface as an open subset, so to answer your question we might as well show that S(Q) is dense in S(R)If x y z = 8, xy yz zx = 15 then find x3 y3 z3 3xyz 1 151 2 152 3 153 4 251 5 180 In this question formula a 3 b 3 c 3 3abc = (abc)(a 2 b 2 c 2 abbcca) is used LHS part (xy) 3 (yz) 3 (zx) 3 3(xy)(yz)(zx) =(xy yz zx
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Verify That X3 Y3 Z3 3xyz 1 2 X Y Z X Y 2 Y Z 2 Z X 2 Class 9th Ex 2 5 Question 12 Youtube
Now the formula x 3 y 3 z 3 – 3xyz = (x y z) (x 2 y 2 z 2 – xy – yz – zx) Putting the given values ⇒ 80 – 3xyz = 8 × (24 – ) ⇒ 3xyz = 80 – 32 = 48 ⇒ xyz = 16 ∴ Required value of xyz = 16 Short trick Putting, x = y = 2 and z = 4Answer (1 of 2) Put x y = a, y z = b and z x = c Then a b c = x y y z z x = 0 Since a³ b³ c³ 3abc = (a b c)(a²b²c² ab bc caX 3 y 3 z 3 − 3 x y z = (x y z) (x 2 y 2 z 2 − x y − y z − x z)
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Note that (can be easily seen with rule of Sarrus)$$ eginvmatrix x & y và z \ z và x và y \ y & z & x \ endvmatrix=x^3y^3z^33xyz$$ On the other hand, it is equal to lớn (if we địa chỉ cửa hàng to the first row 2 other rows)$$ eginvmatrix xyz và xyz & xyz \ z & x và y \ y & z và x \ endvmatrix=(xyz) eginvmatrix 1 & 1 & 1 \ z và x & y \ y & z & x \ endvmatrix=(xyz
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