To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Verify that `x^3y^3z^33x y z=1/2(xyz)(xy)^2(yz)^2(zx)^2` If x, y, z are different and Δ = (x, x2, 1 x3), (y, y2, 1 y3), (z, z2, 1 z3) = 0 then show that 1 xyz = 0 We have Now, we know that If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants Click here 👆 to get an answer to your question ️ Solve for (x, y, z) if x y z = 3, x2 y2 z2 = 3, x3 y3 z3 = 3 krishnapohekar6 krishnapohekar6
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Prove that (x-y)^3 (y-z)^3 (z-x)^3=3(x-y)(y-z)(z-x)
Prove that (x-y)^3 (y-z)^3 (z-x)^3=3(x-y)(y-z)(z-x)- 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 it out toMathcircler "Previous Answer It depends If you want three Real numbers x ,y , z satisfying x^ 3 y^ 3 z ^3 =33 then there are infinitely many, and they are very easy to find If you want three Ra
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Prove that in the expansion of `(1x)^n(1y)^n(1z)^n` , the sum of the coefficients of the terms of degree `ri s^(3n)C_r` A `(""^(n)C_(r))^(3)`Using x = 2, y = 3, and z = 4, evaluate each expression and match to its corresponding answer SW 1 ху 2 A) 6 2 х(у 2) B) 1 3 2z Зу C) 2 4 x yz D) 14 5 ху— хz E) 12 6 2(г— х) F) 1 7 yz X y G) x y 8 What is the value of x 3y z if x = 3, y = 3, and z = 4 1 See answer gheedominique is waiting for your help Add your answer and earn points kphung kphung Substitute the values 33(3)4 First evaluate the multiplication 394 39=12 124=16 Final answer16 New questions in Biology
(((x 3)•(yz))y 3 •(zx))z 3 •(xy) Step 3 Equation at the end of step 3 (x 3 •(yz)y 3 •(zx))z 3 •(xy) Step 4 Trying to factor by pulling out 41 Factoring x 3 yx 3 zxy 3 xz 3 y 3 zyz 3 Thoughtfully split the expression at hand into groups, each group having two terms Group 1 y 3 zxy 3 Group 2 x 3 yx 3 zSolve for z 2/x=3/y1/z 2 x = 3 y 1 z 2 x = 3 y 1 z Rewrite the equation as 3 y 1 z = 2 x 3 y 1 z = 2 x 3 y 1 z = 2 x 3 y 1 z = 2 x Subtract 3 y 3 y from both sides of the equation 1 z = 2 x − 3 y 1 z = 2 x 3 y Find the LCD of the terms in the equation Tap for more steps` X ` Y Z =` 33X ` ` ` Z = 5 ` X 2Y Z =` 3 ` Here are the seven steps 1 Pick a letter (If one of the equations has only two letters, choose the ` `letter that is missing) 2 Pick an equation that contains that letter 3 Solve for that letter in that equation 4 Substitute what you get into any other equations that contains that
2^x 3^y = z^2 1 where x, y, and z are nonnegative integers We can see that 1 is a congruence modulo 8 3^y = z^2 (mod 8) 2 When z=1 and and y=2 we have 3^2 = 1 (mod 8) 3 and by Fermat's theorem So this means that y must be even As y is even, we can write y = 2k for some integer k Move all terms containing x to the left, all other terms to the right Add '1yz' to each side of the equation xy xz yz 1yz = t 1yz Combine like terms yz 1yz = 0 xy xz 0 = t 1yz xy xz = t 1yz Reorder the terms 1t xy xz yz = t 1yz 1t yz Reorder the terms 1t xy xz yz = t 1t 1yz yz CombineUsing Formula, a^3 b^3 c^3 3abc = (a b c)(a^2 b^2 c^2 ab ac bc) Adding 3abc both side a^3 b^3 c^3 = (a b c)(a^2 b^2 c^2 ab ac bc
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Click here 👆 to get an answer to your question ️ prove that (xy)³(yz)³(zx)³3(xy)(yz)(zy)=2(x³y³z³3xyz) The answer is z=3(axy) Beginning Equation a=xyz/3 First multiply both sides of the equation by 3 3a=3x3yz Subract 3x and 3y from both sides 3a3x3y=z Factor out the 3 3(axy)=z Show that $$(xy)^3(yz)^3(zx)^3 = 3(xy)(yz)(zx)$$ This can be shown through expansion but there is a more elegant solution I cannot discover anything I would consider elegant Can anyone h
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To evaluate the expression, substitute the given values for x, y and z into it The expression can be simplified by collecting like terms #rArr2x^32y^2z^4#3*3x2*y*4z*x*y*z This deals with simplification or other simple results Overview Steps Topics Terms and topics Links Related links 1 solution (s) found (3^2*2^2x^3y^2z^2)Click here👆to get an answer to your question ️ If x y z = 0 , then x^3 y^3 z^3 =
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(xy)^3 (yz)3 (zx)^3 >= 3(xy)(yz)(zx) But that is not the question set Please note that this is the first chapter and all that has been covered is basic number theory, rational powers, inequalities and divisibility I am assuming those are the only tools I have at my disposal, I have not been introduced to any identities etcY=xz/3 Simple and best practice solution for y=xz/3 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 prove that (xy)3(yz)3(zx)33(xy) (yz) (zx) =2(x3y3 z 3 3xyz) Maths Polynomials Using identity a 3 b 3 c 3 3abc = (abc)(a 2 b 2 c 2 ab
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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 ____No integers x;y;z with xyz6= 0 satisfy x3 y3 z3 = 0 Proof We may assume that x, y, and zare pairwise coprime If xyzis not divisible by 3, then the equation has no solution even in Z=(9), where every nonzero cube is 1 Suppose then, without loss of generality, that 3jz We will work in the UFD R= Z with = ( 1i p 3)=2, a root of theThere are infinite solutions as you have 3 variables but only one equation However, it is interesting to note that there is no solution for which all of the varibles x,y,z are integersFermat's last theorem says that there is no solution of x^n
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