what is the sum of all the coefficients in the expansion of (x+3y-z)^25?
913842
25x+75y-25z\nSum of all coefficients\n= 25+75-25\n=75
660602
d/dx[y/x^2] gives y'/x^2 - 2y/x^3 on expansion
983933
(x+y+z)(x+y+z) = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\n\n(1-x^2 - y^2 - z^2) = 2xy + 2xz + 2yz\n\n(1 - x^2 - y^2 - z^2) \n------------------------ = xy + xz + yz\n.............2\n\nClearly the left hand side is less than 1/2
340490
(ax^2 + bx + c)(dx^2 + ex + f)\n=ax^2(dx^2 + ex + f) + bx(dx^2 + ex + f) + c(dx^2 + ex + f)\n=adx^4 + aex^3 + afx^2 + bdx^3 + bex^2 + bfx + cdx^2 + cex + f\n=adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + f
1058665
|sin(z)|^2 + |cos(z)|^2\n\nWhere |f(z)|^2 is understood to mean f(z)*~f(z), where ~f(z) is the complex congugate of f(z): ~f(x + iy) = f(x - iy)\n\nWriting the real and imaginary parts of the arguments explicitly, we have:\n\n= |sin(x+iy)|^2 + |cos(x+iy)|^2\n\nUsing the sum of angles formulas:\n\n= |(sin(x)cos(iy) + cos(x)sin(iy)|^2 + \n  |(cos(x)cos(iy) - sin(x)sin(iy)|^2\n\nExpanding in terms of complex congugates:\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(-iy) + cos(x)sin(-iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(-iy) - sin(x)sin(-iy))\n\nNoting that cos(-x) = cos(x) and sin(-x) = -sin(x):\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(iy) - cos(x)sin(iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(iy) + sin(x)sin(iy))\n\n\nMultiplying this out (both terms are the product of a sum and difference of two terms, which yield the difference of squares):\n\n= (sin(x)cos(iy))^2 - (cos(x)sin(iy))^2 + (cos(x)cos(iy))^2 - (sin(x)sin(iy))^2\n\nFactoring:\n\n= ((sin(x))^2 + (cos(x))^2)*cos(iy) - ((sin(x))^2 + (cos(x))^2)*sin(iy)\n\nbut ((sin(x))^2 + (cos(x))^2) = 1, so\n\n= (cos(iy))^2 - (sin(iy))^2\n\nFrom Euler's formula (exp(ix) = cos(x) + i*sin(x)), we can show that:\n\ncos(iy) = (exp(-y) + exp(y))/2 = cosh(y)\nand\nsin(iy) = (exp(-y) - exp(y))/(2*i) = i*sinh(y)\n\nSubstituting these identities in for cos(iy) and sin(iy), we have:\n\n= (cosh(y))^2 - (i*sinh(y))^2 \n\n= (cosh(y))^2 + (sinh(y))^2\nQED
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
4959
=8(y^3)z + 12(y^2)(z^2) - 20y^2
739976
Add all the polynomials together and you get:\nx^4 y^4
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
250266
The fourth term in a sixth order binomial expansion has a coefficient given by 6 choose 3. This is because we need to consider all rows of Pascal's triangle beginning with row zero, column zero. 6 choose 3 falls on row 6 column 3 and is 6!/(3!*3!)=20. This factor will then be multiplied by x^3 as well as (-2y)^3. Together the fourth term will be:\n\n-160x^3y^3
395170
Did you mean (3x^7)(x^-1)(y^5)/(6x^3z)? or what you put up as your subject line?\n\nIf the one in your subject line is correct then:\n\nFirst step: Simplify\n(x^7x^-1y^5)/(2x^3z)\n\nSecond: Add together all the like terms.\n(x^6y^5)/(2x^3z)\n\nThird: Subtract the number of x's in the numerator by the x's in the denominator.\n\nx^6-x^3=x^3\n\nAnswer= x^3y^5/2z
952689
5x+3y+z+5w=-2\n3x+2y+2z+4w=2\n2x+4y+3z-3w=-11\n4x-3y-2z+2w=-3\n\n=>\n5     3     1     5   I  -2\n3     2     2     4   I   2\n2     4     3    -3   I  -11 \n4    -3    -2     2   I  -3\n\nR4 - (4/5)*R1 , R3 - (2/5)*R1 and R2 - (3/5)*R1 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    2.8000    2.6000   -5.0000  -10.2000\n         0   -5.4000   -2.8000   -2.0000   -1.4000\n\nR4 - (-5.4/0.2)*R2 , R3 - (-5.2/0.2)*R2 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000   35.0000   25.0000   85.0000\n\nR4 - (-37/17)R3 will give\n\n5.0000   3.0000    1.0000    5.0000   -2.0000\n         0   0.2000    1.4000    1.0000    3.2000\n         0   0.0000  -17.0000  -19.0000  -55.0000\n         0   0.0000    0.0000  -14.1176  -28.2353\n\nR4 / -14.1176 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 - (-19)R4 will give\n\n    5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0   -0.0000  -17.0000         0  -17.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR3 /(-17) will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000    1.0000    0.0000    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- R4 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000             0    1.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- (1.4)R3 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    0.0000             0   -0.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 /(0.2) will give \n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR1- 5R4 will give\n\n5.0000    3.0000    1.0000             0  -12.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-R3 will give \n\n5.0000    3.0000             0             0  -13.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-3R2 will give \n\n5.0000             0   -0.0000             0  -10.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1/5 will give \n1.0000             0   -0.0000             0   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nAnd thats the final answer.\n\nSo, your answer is x=-2, y=-1,z=1, w=2\n\nVerification\na=\n     5     3     1     5\n     3     2     2     4\n     2     4     3    -3\n     4    -3    -2     2\nb= \n    -2\n     2\n   -11\n    -3\ninverse of a is\n\n    0.0250   -0.0250    0.1250    0.1750\n    0.4000   -0.4000    0.0000   -0.2000\n   -0.5125    0.6792    0.1042    0.0792\n    0.0375    0.1292   -0.1458   -0.0708\n\nThus,\ninv(a)*b = [-2 -1 1 2]'\n\nI hope i answered your question.\n\nAll the best
913862
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
220478
(x^2y - xy^2) = xy*(x-y) = 7*5 = 35
669578
I have corrected the expression. Well here I redo the factorization\n\nThe given expresion \n= 12x^3+60x^2y+75xy^2\n= 3x(4x^2+20xy+25y^2)\n= 3x[(2x)^2+2(2x)(5y)+(5y)^2]\n= 3x(2x+5y)^2
947283
You can do this using "synthetic division"\n....... ______________________\nx - 5 ) x^4 - 0x^3 - 3x^2 + 0x + 6\n\nI'm going to simplify this leaving out the x-terms and only showing the coefficients, but just imagine they are there:\n\n.............. 1 . 5 .22 110\n1 -5 ) 1 . 0 .-3 . 0 . 6\n......... 1. -5\n......... ------\n.............. 5 .-3\n.............. 5 -25\n.............. -------\n................... 22. 0\n....... ........... 22 -110\n....... ............----------\n....... ........ ........ 110  6\n....... ........ ........ 110 -550\n....... ........ ........ ------------\n....... ........ ........ ....... 556\n\nSo if I did it correctly the answer should be:\n1 5 22 110 remainder 556\n\nThis equates to:\n(x^3 + 5x^2 + 22x + 110) + 556/(x-5)
662979
If you expand (x+y)^n in terms of Binomial Theorem, you will see that\n\n(x+y)^n = C(n,0)*(x^n) + C(n,1)*(x^n-1)*y + ... + C(n,n)*(y^n)\n\nThe coefficients of this expansion are the C's which represent Combinations of values. Hence, sum of the co-efficients equals:\n\nC(n,0)+C(n,1)+...C(n,n) = 4096.\n\nUsing an simple mathematical proof, we can show that the sum of the above combinations equals 2^n. Thus,\n\n2^n = 4096\n\nSolving for n, we get:\n\nn = 12\n\n\nNow, to find the greatest cofficient of expansion, we look into all the coefficients together:\n\nC(12,0) = 1\nC(12,1) = 12\nC(12,2) = 66\n...\n...\nC(12,10) = 66\nC(12,11) = 12\nC(12,12) = 1\n\nYou will find a symmetry when you calculate the coefficients & you can easily judge that the maximum value lies exactly in the middle of the list, i.e. C(12,6)\n\nThus the greatest cofficient in the expansion is:\n\nC(12,6) = (12*11*10*9*8*7)/(1*2*3*4*5*6) = 924\n\nThus the greatest coefficient of expansio for your equation is 924 - which turns out to be the coefficient of (x^6)*(y^6)\n\n\n\n\n\nP.S. Oops... I feel kinda ashamed that I said 'advanced mathematical proof - it is actually a very simple proof as what 'geek24' said.