Elaborating x 3 y 3 using identity a 3 b 3 = (a b)(a 2 ab b 2) = x ( x y)(x 2 xy y 2) 3xy (x y ) Taking common x( x y ) in both the terms = x ( x y){x 2 xy y 2 3y} ∴ x (x 3 y 3) 3xy ( x y) = x ( x y )(x 2 xy y 2 3y)See the answer See the answer See the answer done loading Show transcribed image text Expert Answer Who are the experts?Factorise (X Y)3 8x3 CISCE ICSE Class 9 Question Papers 10 Textbook Solutions Important Solutions 5 Question Bank Solutions Concept Notes & Videos 242 Syllabus Advertisement Remove all ads Factorise (X Y)3 8x3 Mathematics
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Factorise x(x^3-y^3)+3xy(x-y)- Explanation differentiate implicitly with respect to x The product rule has to be used on the right side 3x2 3y2 dy dx = 3x dy dx 3y ⇒ 3y2 dy dx −3x dy dx = 3y − 3x2 ⇒ dy dx (3y2 −3x) = 3(y − x2) ⇒ dy dx = 3(y − x2) 3(y2 − x) = y − x2 y2 − x Answer linkFactor out the Greatest Common Factor (GCF), '4dxy' 4dxy(y 2 1x 2) = 0 Factor a difference between two squares 4dxy((y x)(y 1x)) = 0 Ignore the factor 4 Subproblem 1 Set the factor 'dxy' equal to zero and attempt to solve Simplifying dxy = 0 Solving dxy = 0 Move all terms containing d to the left, all other terms to the right
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How do you factor completely x^3 y^3 z^3 3xyz?For xaxis to be the tangent of the curve x 3 y 3 = 3 x y, first you need to check if the curve intersects or touches with xaxis Since on xaxis y = 0 which should also satisfy x 3 y 3 = 3 x y Substituting y = 0, we obtain x 3 0 = 0 Which is satisfied for x = 0 Hence the curve touches or intersects with xaxis at (0, 0)Differentiate with respect to x 3x^2 3y^2dy/dx=3(xdy/dx y) 3x^23y^2dy/dx =3xdy/dx3y 3y^2dy/dx3xdy/dx=3y3x^2 dy/dx(3y^2–3x)=3y3x^2 dy/dx=(3y3x^2)/(3y^2–3x
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(a) x 2 y 2 2xy (b) x 2 y 2 – xy (c) xy 2 (d) 3xy polynomialsAlgebra 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 (x−y)(x2 xyy2) ( x y) ( x 2 x y y 2)Click here👆to get an answer to your question ️ Factorise x^3 2x^2y 3xy^2 6y^3 Join / Login maths Factorise x 3 − 2 x 2 y 3 x y 2 − 6 y 3 Answer Given x 3 − 2 x 2 y 3 x y 2
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(x 3 y 3) = (x y)(x 2 xy y 2) We can factor the whole thing out as x(x y)(x 2 xy y 2) 3xy(x y) The common factor of the two terms is x(x y)We think you wrote (x^(2)xzxyyz)/(x^(2)y^(2))(xz)/(x^(3)y^(3)) This deals with factoring binomials as the sum or difference of cubesLim (x, y) = (1,2) 4x2 y2 O a 2 3 Ob nm mln O ო და 1 OC 9 d does not exist O e 4 3 3 4 o g 4 9 This problem has been solved!
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0 This is an online question, and the system is marking it to be incorrect However, I can not figure out where I went wrong Calculate the derivative of y with respect to x x 3 y 3 x y 3 = x y Here is my attempt ( x 3 y) ′ ( 3 x y 3) ′ = x ′ y ′ ( 3 x 2 y d y d x x 3) 3 ( y 3 3 y 2 d y d x x) = 1 d y d x 3 x 2 yThat is (a^3b^3c^3)= (abc) (a^2b^2c^2abbccb)3abc From the above identity here a=3x b=y c=z expand according to identity ,we get (3xyz) (9x^2y^2z^23xyyzyz33xyz) (3xyz) (9x^2y^2z^23xy3xz3xz)33xyz9xyz3xy 2 3xy 2 • x 3 3xy 2 = ———— = ————————— 1 x 3 Checking for a perfect cube x 6 3x 5 y 3x 4 y 2 y 3 is not a perfect cube
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Transcript Ex 25, 9 Verify (i) x3 y3 = (x y) (x2 – xy y2) LHS x3 y3 We know (x y)3 = x3 y3 3xy (x y) So, x3 y3 = (x y)3 – 3xy (x yWelcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get= x(x – y) (x^2 xy y^2 3y) = x(x – y) (x^2 xy y^2 3y)
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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 যদি `A= (x^3y^3)/ ( (xy)^23xy), B= ( (xy)^23xy)/ (x^3y^3), C= (xy)/ 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 Errorless Vol1 Errorless Vol2 Maths The function F(x, y) = x 3 y 3 3xy 4 First order partial derivatives f x (x, y) = 3x 2 3y f y (x, y) = 3y 2 3x Second order partial derivatives F xx (x, y) = 6x F yy (x, y) = 6y F xy (x, y) = 3 Step 2 The critical points satisfy the equations f x (x,y) = 0 and f y (x,y) = 0 So solve the following equations F x = 0 and F y
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Click here👆to get an answer to your question ️ factorise x^3 3x^2y 3xy^2 y^3 8FRANK Solutions Class 9 Maths Chapter 5 Factorisation 1 Factorise the following by taking out the common factors (a) 4x2y3 3– 6x y2 – 12xy2 (b) 5a (x 2 – y 2) 35b (x – y ) (c) 2x5y 8x3y2 2– 12x y3 (d) 12a3 2 15a b – 21ab2 (e) 24m4n6 6 56m n4 – 72m2n2 (f) (a – b)2 – 2(a – b) (g) 2a (p 2 q2) 4b (p q2) (h) 81 (p q)2 – 9p – 9q (i) (mx ny)2 2 (nx – my) 10 A polynomial from Qx, y, z is a polynomial from Qx, yz, so it can be viewed as a polynomial in z with coefficients from the integral domain Qx, y p(z) = z3 − 3xy ⋅ z x3 y3 So we can try our methods to factor a polynomial of degree 3 over an integral domain If it can be factored then there is a factor of degree 1, we call
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X^33x^2y3xy^2y^3 ———————————— X^3y^3 Get the answers you need, now!Get the answer to this question and access a vast question bank that is tailored for studentsSimple and best practice solution for (xy^3y^2senx)dx= (3xy^22ycosx)dy equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it
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Question factor x^3y^3 into irreducibles in Q(x,y) and prove that each of the factors is irreducible Answer by richard1234(7193) (Show Source) You can put this solution on YOUR website! 1 Follow 0 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 =0Question The equation x3 3xy y3 = 1 is solved in integers Find the possible values of xy Found 3 solutions by Alan3354, Edwin McCravy, richard1234
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(x•((x 3)(y 3)))3xy•(xy) Step 2 Trying to factor as a Difference of Cubes 21 Factoring x 3y 3 Theory A difference of two perfect cubes, a 3 b 3 can be factored into (ab) • (a 2 ab b 2) Proof (ab)•(a 2 abb 2) = a 3 a 2 b ab 2ba 2b 2 ab 3 = a 3 (a 2 bba 2)(ab 2b 2 a)b 3 = a 3 0 0b 3 = a 3b 3 Check x 3 is the cube of x 1If xy=5 and xy=6 then find x^3y^3 xy=5 and xy=6y=6x If 52x1 25x1 = 2500 then the value of x = Consider If 2x5 is a factor of 2x^25 find the value of x if 2Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
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Share Consider x^ {2}3xy2y^ {2}4x7y3 as a polynomial over variable x Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {2} and m divides the constant factor 2y^ {2}7y3 One such factor is xy3 Factor the polynomial by dividing it by this factorSolve 3xy′−12y=x−1,y(1)=−8 3 x y ′ − 12 y = x − 1 , y ( 1 ) = − 8 (a) Identify the integrating factor, α(x) α ( x ) α(x) α ( x ) = (b) Find the general solution y(x)= y ( xWhat is the value of (1)^1 Number name international place value chart for the number
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Given the following function f (x, y) = 3xy 3x x^3 3y^3 a) Evaluate analytically partialdifferential f/partialdifferential x, partialdifferential f/partialdifferential y, and partialdifferential^2 f/ (partialdifferential x partialdifferential y) at x = y = 1 b) Evaluate numerically the same terms given in a) with delta x = delta y = 0 This is a semiimportant identity to know (x^3y^3)=(xy)(x^2xyy^2) Although it doesn't apply directly to this question, it's also important to know that (x^3y^3)=(xy)(x^2xyy^2) This gives us the rule (x^3y^3)=(xy)(x^2∓xyy^2) Which of the following is a factor of (x y) 3 – (x 3 y 3 )?
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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 WolframAlpha brings expertlevel knowledge andHello students, today we are going to learn algebraic identities (XY)^3 = X^3 3XY (XY) Y^3 (XY)^3 = X^3 3XY (XY) Y^3#algebraicidentities #maX^3 y^3 = 3xy^2, find dy/dx using implicit differentiation
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`therefore 3xy` is a factor of `(xy)^3(x^3y^3)` ← Prev Question Next Question → Related questions 0 votes 1 answer If `(x1)` is a factor of the polynomial `Question 11 – Factorise `27x^3 y^3 z^3 9xy\z` Answer Given; Factorize x(x^3y^3) 3xy (xy) asked in Mathematics by Sonu khan (356k points) factorisation of algebraic expressions;
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`27^3 y^3 z^3 9xy\z` `= (3x)^3 y^3 z^3 3xx3xy\z` Using the identity `x^3 y^3 z^3 3xy\z` `= (x y z)(x^2 y^2 z^2 xy – yz – xz)` We get `(3x y z)(3x)^2 y^2 z^2 3xy – yz – 3xz` `= (3x y z)(9x^2 y^2 z^3 3xy – yz – 3xz)`
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