++ 50 ++ (x-y)^3=x^3-y^3-3xy(x-y) 351854-X(x^3-y^3)+3xy(x-y) solution

So we set it = 0 i*sqrt(3)(x1) = 0 x1 = 0 x = 1 Then substituting that y = y = y = y = 1 So we end up with all solutions {(x,y) x y = 1} plus the one solution (x,y) = (1,1) Now if you were asking about xy instead of xy, the answer would be xy is always either 1 or 2 Are you sure you didn't make a typo and you were asking about xy and not xy? = x^3 3x^2y 3xy^2 y^3 = x^3 y^3 3xy(x y) Also, Read Cube of a Binomial Cube of Sum of Two Binomials Examples 1 Determine the expansion of (x 2y)^3 Solution The given expression is (x 2y)^3 We have an equation on cubes like (x y)^3 = x^3 y^3 3xy(x y) By comparing the above expression with the (x y)^3 Here, xWe know that (x y) 3 = x 3 y 3 3xy (x y) Using Identity VII ⇒ x 3 y 3 = (x y) 3 3xy (x y) x 3 y 3 = (x y) { (x y) 2 3xy} ⇒ x 3 y 3 = (x y) (x 2 2xy y 2 3ry) Using Identity IV ⇒ x 3 y 3 = (x y) (x 2 xy y 2 ) 350 Views

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X(x^3-y^3)+3xy(x-y) solution

X(x^3-y^3)+3xy(x-y) solution-Ĺet x^yy^x= (xy)^3=x^3y^33xy (xy) Or,x^yy^x=x^3y^39xy again x=3y so x^3y^39xy= (3y)^3y^39 (3y)y=2727y9y^2y^3y^39y^227=5427y Now we got 54 27y=27, or 27y=27,y=1 Then x=31=2 The values of x& y only satisfy xy=3 and doen't satisfy x^yy^x=27 So there is no solution for the values of x & y kash EduTech Pvt Ltd What are you looking for?

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 If x y = 12, and xy = 27, then find the value of x3 y3 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queriesĹet x^yy^x=(xy)^3=x^3y^33xy(xy) Or,x^yy^x=x^3y^39xy again x=3y so x^3y^39xy=(3y)^3y^39(3y)y=2727y9y^2y^3y^39y^227=5427y Now we got 54 27y=27, or 27y=27,y=1Polynomial Examples Find the remainder when x 4 x 3 – 2x 2 x 1 is divided by x – 1 Solution Here, p(x) = x 4 x 3 – 2x 2 x

X^3 y^3 = (x y)(x^2 xy y^2) (identity) (2i) = (2i)(x^2 xy y^2) Divide both sides by 2i x^2 xy y^2 = 1 Add 3xy to both sides x^2 2xy y^2 = 3xy 1 (x y)^2 = 3xy 1 (2i)^2 = 3xy 1 4 = 3xy 1 3xy = 3 xy = 1 y = 1/x, xX 0 =1 x a y a = (xy) a, 2 2 3 2 = 6 2;(xy) 3 = x 3 y 3 3xy(x y) x 2 y 2 = (x y)(x y) x2 = 1/x 2, 24 = 1/16 = 1/2 4 (x a)(x b) = x ab;

It is clear that when $x=y$ we have $x^3y^3=0$ Then use long division to divide $x^3y^3$ by $xy$ and the result will be the equation on the right Another way would be to write $$\left(\frac{x}{y}\right)^3 1$$ Now we wish to find the zeros of this polynomialPhilip Macdonald Answered 1 year ago Author has 234 answers and 413K answer views xy =7, xy =1 (xy)^3 = x^3 3x^2*y 3x*y^2 y^3 = x^3y^3 3xy (xy) Then 7^3 = x^3y^3 3*1*7, ie, x^3y^3 = 7^3 7*3 = 7 (7^2 3) = 7*46 =322 143 views Sponsored by Jumbo Privacy & Security (x y z) 2 = x 2 y 2 z 2 2xy 2yz 2zx (x y) 3 = x 3 y 3 3xy(x y) (x – y) 3 = x 3 – y 3 – 3xy(x – y) x 3 y 3 z 3 – 3xyz = (x y z)(x 2 y 2 z 2 – xy – yz – zx;

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Solution (By Examveda Team) Given, xy = 2 cubing both sides (xy) 3 = 2 3 => x 3 y 3 3xy ( xy) = 8 => x 3 y 3 3×15×2= 8 => x 3 y 3 90 = 8 => x 3 y 3 = 035 (1) Upvote (1) Choose An Option That Best Describes Your Problem Answer not in Detail Incomplete Answer Answer Incorrect Others Answer not in Detail Incomplete Answer Answer IncorrectCube of a binomial can be simplified using the identities \({(x y)}^3 = x^3 y^3 3xy(x y)\)

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 4 It's two equations, with two variables We can do what we always do You can write y = 8 / x, and substitute into the first equation, obtaining x3 / x3 = 72 From here, we want to write this as a polynomial, so we multiply through x3 so there are no negative powers, and then move everything to the left(xy)^3 x^3y^33xy(xy) x^3y^3z^33xyz (xyz)(x^2y^2z^2xyyzzx) √ab √a√b √a/b(whole) √a/√b (√a√b)(√a√b) ab (a√b)(a√b) a^2b (√a√b)^2 a2√abb OTHER SETS BY THIS CREATOR GeographyChapter 1Indiasize and location 13 terms ikeepchangingmynameStudy on the go Download the iOS Download the Android app Company About Us Scholarships

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 kdtiwari9862 is waiting for your help Add your answer and earn points ashwinipmashwinipm Answer 3xyz Stepbystep explanation x y z = 0 = x y = z Cubing on both sides (x y)^3 = (z)^3 ⇒x ^3y ^33xy(xy)=z ^3Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyX 3 y 3 3xy = 1

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 Here, exponent of every variable is a whole number, but x 10 y 3 t 50 is a polynomial in x, y and t, ie, in three variables So, it is not a polynomial in one variable Ex 21 Class 9 Maths Question 2 (ii) (2a – 3b) 3 = (2a) 3 – (3b) 33(2a)(3b)(2a3b) Using identity (xy) 3 =x 3y 33xy(xy) = 8a 327b 318ab(2a3b) = 8a 3 – 27 b 3 – 36a 2 b 54ab 2 =8a 3 – 36a 2 b 54ab 2 – 27 b 3 Question 7 Evaluate the following using suitable identities (i) (99) 3Quadratic equation, for any given x if ax 2 bx c =0 then x has 2 solutions x=(b√(b 2 4ac)/2a, x=(b√(b 2 4ac)/2a x a y b is not equal to (xy) ab

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 0 Follow 0 A K Daya Sir, added an answer, on 25/9/13 A K Daya Sir answered this x 3 y 3 = (x y) (x 2 xy y 2 ) this formula can be derived from (x y) 3 = x 3 y 3 3xy (x y) x 3 y 3 = (x y) 3 3xy (x y) x 3 y 3 = (x y) (x y) 2 3xy = (x y) x 2 y 2 2xy 3xy = (x y) (x 2 xy y 2 ) Was this answer helpful?Simplify (xy)^3 (x − y)3 ( x y) 3 Use the Binomial Theorem x3 3x2(−y) 3x(−y)2 (−y)3 x 3 3 x 2 ( y) 3 x ( y) 2 ( y) 3 Simplify each term Tap for more steps Rewrite using the commutative property of multiplication x 3 3 ⋅ − 1 ( x 2 y) 3 x ( − y) 2 ( − y) 3 x 3 3 ⋅ 1 ( x 2 y) 3 x ( y) 2 find x^3y^3 The coordinates of point Y are giving The midpoint XY is (3,5) Find the coordinates of point X

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Typo/misspeak around 4 minutes 369=27Multivariable Calculus Find all local maxima/minima and saddle points for the function f(x,y) = x^3 3xy y^3 WHere given are, XY=7, let's say it is formula 1 in order to make it kinda convenient and easy to understand And XY=10, let's mark it as 2 Now, see formula 1 can be also written as,1 How do you simplify the cube of a binomial?

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 Which of the following is a factor of `(xy)^3(x^3y^3)?` A `x^2y^22xy` B `x^2y^2xy` C `xy^2` D `3xy` Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queriesSOLUTION ⇒ x – y = 3 ⇒ (x – y) 3 = 3 3 ⇒ x 3 – y 3 – 3xy (x – y) = 27 ⇒ x 3 – y 3 – 3xy × 3 = 27 ⇒ x 3 – y 3 – 9xy = 27(i) (99) 3 (ii) (102) 3 (iii) (998) 3 Solution We will use the following algebraic Identities to evaluate the given expressions (x y) 3 = x 3 y 3 3xy(x y) (x y) 3 = x 3 y 3 3xy(x y) (i) (99) 3 = (100 1) 3 Identity Here we can use (x y) 3 = x 3 y 3 3xy(x y) Let us substitute x = 100, y = 1 (99) 3 = (100) 3 (1) 3 3(100)(1)(100 1)

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 Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given (i) Area 25a2 – 35a 12 (ii) Area 35y2 13y – 12 Solution (i) We have, area of rectangle = 25a 2 – 35a12 = 25a 2 – a – 15a12 Thanks for trying but the answer is 459 dududsu2 #3 1 x= 9y xy would then be (9y) y = 10 y^2 9y 10 = 0 y^29y10 = 0 y = 9/2 sqrt41/ 2 (x^3 y^3 = ( Start your 48hour free trial to unlock this answer and thousands more Enjoy eNotes adfree and cancel anytime

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Get FREE NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 25 We have created Step by Step solutions for Class 9 maths to help you to revise(x y)^3 = (x^3) (y^3) 3xy(x y) so, (3)^3 = 9 3xy(3) so, 27 = 9 9xy so, (27 9)/9 = xy so, xy = 18/9 = 2X a /x b = x ab = 1/x ba;

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 = (4a – 3b) 3 ∵ (x – y) 3 = x 3 – y 3 – 3xy(x – y) = (40 – 3b) (4a – 3b) (4a – 3b) Question 3 What are the possible expressions for the dimensions of a cuboid whose volume is given below ?यदि x 2 – 9x 1 = 0 है, तो x 3 1/x 3 का मान क्या है?In mathematics, the cube of sum of two terms is expressed as the cube of binomial x y It is read as x plus y whole cube It is mainly used in mathematics as a formula for expanding cube of sum of any two terms in their terms ( x y) 3 = x 3 y 3 3 x 2 y 3 x y 2

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Volume = 12ky 2 8ky – k Solution We have, volume = 12ky 2 8ky – k = 4k(3y 2 2y – 5) = 4k(3y 2 5y – 3y – 5)(X – Y) 3 = X 3 – Y 3 – 3XY(X – Y) Or (X – Y) 3 = X 3 – Y 3 – 3X 2 Y 3XY 2 Was this answer helpful?Ĺet x^yy^x=(xy)^3=x^3y^33xy(xy) Or,x^yy^x=x^3y^39xy again x=3y so x^3y^39xy=(3y)^3y^39(3y)y=2727y9y^2y^3y^39y^227=5427y Now we got 54 27y=27, or 27y=27,y=1

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 Heya mate, Here is ur answer xyz=0 xy=z cubing both the sides (xy)^3=z^3 x^3 y^3 3xy (xy) =z^3 but xy=z x^3 y^3 3xy (z) =z^3 x^3 Y^3 3xyz=z^31 Explanation We know that algebraic formula, (x y) 3 = x 3 y 3 3xy (x y) put the value of x y in given equation given, x y = 1 1 = x 3 y 3 3xy X 1?Solve 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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 xy=1 (xy)³=1 x³3x²y3xy²y³=1 3xy(xy)3=1 xy=2/3 (xy)²=1 x²y²2xy=1 x²y²=7/3 (x²y²)(x³y³)=7/3*3=7 x^5x²y³x³y²y^5=7 x^5y^5=x²y²(xy)7=4/97=59/9 已赞Telangana SCERT Class 9 Math Chapter 2 Polynomials and Factorisation Exercise 25 Math Problems and Solution Here in this Post Telanagana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 25Use identity ( a b)^3 = a^3 b^3 3ab (a b ) Put a= x and b= y ( x y)^3 = x^3 y^3 3xy ( x y ) In further step 3xy can be multiplied inside the bracket The answer is 👍 Helpful 💡

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Note that (x y) 3 = x 3 y 3 3xy(x y) Share this link with a friend Copied! 展开全部 (xy)^3= (xy)^2* (xy)= (x^22xyy^2)* (xy)=x^3y^33xy (xy) 所以得到上述证明 这是一个简单的证明，你应该还没有接触到 (xy)^3 4 已赞过 已踩过 < 你对这个回答的评价是？ 分享 新浪微博 Q9 Verify (i) x 3 y 3 = (x y) (x 2 – xy y 2) (ii) x 3 – y 3 = (x – y) (x 2 xy y 2) Answer (i) x 3 y 3 = (x y) (x 2 – xy y 2) ∵ (x y) 3

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SOLUTION ⇒ x – y = 3 ⇒ (x – y) 3 = 3 3 ⇒ x 3 – y 3 – 3xy (x – y) = 27 ⇒ x 3 – y 3 – 3xy × 3 = 27 ⇒ x 3 – y 3 – 9xy = 27 Related Questions If x 2 – 9x 1 = 0, then what is the value of x 3 1/x 3?

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Incoming Term: (x+y)^3=x^3+y^3+3xy(x+y), x(x^3-y^3)+3xy(x-y) solution, x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3, (x+y)3=x3+y3+3xy(x+y),