Can this  equation X^n + Y^n + Z ^n = R^n,be transfromed into an equivalent geometrical structure?
101215
Obviously n=2 is a sphere, and n=1 is plane at 45degrees, and n=0 is a defective case.\n\nYou're talking about some geometric generalization of a sphere, with a non-Cartesian distance function (X^n + Y^n + Z ^n).\n\nFor general integer n>=3, or for non-integer n, I can't see a nice geometric interpretation, other than this:\nIf you set n~2 like say n = 2.1 or 1.9 you get a distorted sphere. For n>2, the sphere will bulge inwards in the 8 regions where X,Y,Z are not close to 0, for n<2 it bulges outwards.\n\nTry plotting it in Matlab or Octave and see.\n\nThings that this is *not*:\n- an n-dimensional hypersphere with n=3 (x^2 + y^2 + z^2 + ...x_n^2 = r^2), since the exponent is always 2\n- a spheroid\n\n\n\nThis is _remotely_ similar to visualizing Diophantine equations or Fermat's Last Theorem, but allowing X,Y,Z to be reals, not just integers, i.e. (X,Y,Z) not restricted to lattice points (also Fermat only had X,Y,R-terms, no Z-term, but similar)?\nVery interesting question, not aware if people have got anywhere.\n\nGood luck and let us know what you find.
168995
plynomial expression could be factorized as the product of \n(x-r) where r is a root (a zero) of the polynomial soyou must get the roots first\nMany numerical methods exists
220478
(x^2y - xy^2) = xy*(x-y) = 7*5 = 35
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
1124422
[(y^2 + y )/(y+ 2)] / [(y^2 + 3y)/(y^2-4)]\n= [(y^2 + y )/(y+ 2)] times (y^2-4)/(y^2+3y)\nFactor everything\ny^2 + y = y(y+1)\ny^2 + 3y = y(y+3)\ny^2 - 4 = (y+2)(y-2)\nCancel anything on the top with anything on the bottom\nAnswer: (y+1)(y-2)/(y+3)
800132
how does this fit into mathematics?
178683
yeah, I suppose you want me to factorise x^n - y^n. the general formula is\n\nx^n - y^n = (x-y)(x^n-1.y + x^n-2.y^2 + x^n3.y^3+.........+x^2.y^n-2 + x^n-1.y^1)\n\nTry putting n=2 or 3 or4 in this expression and you'll see what I mean.\n\nYou'll arrive at nothing but\n\nx^2 - y^2 = (x-y)(x+y)\nand so on..\n\nAnd please, the next time, a question without exponents. i'm SICK of '^'. \n\nAnd you know what, mr. 'timmobilelaptop' , the next time try using your brain. It's better than any T-189 calculator.
36273
(A^3 + B^3)=(A+B)(A^2 - AB + B^2)
847432
Of course you can... if a = 0, you have an equation that would simplify to:\n\nby = c\ny = c/b  (and since c and b are constants)\ny = d\n\nwhich would represent a horizontal line crossing the y axis at d\n\n\nSame with b, only if b = 0 you would have a horizontal line
7772
You can.  Use the mapping (x, y, z) -> x.  Of course, that is a many-to-one mapping, and kind of useless.
899014
an expression of the form x^-n can be written as \n    1\n_____\n \n   x^n\n\nwhere ^ implies raised to the power of\n\nSo first square the expression you have given. this gives\n\n  4 x^6 y^-6\n__________\n    \n    9 a^12\n\n\ntake reciprocal of this\n\n      9 a^12\n___________\n\n   4 x^6 y^-6\n\nThis is your answer
183757
Assuming you mean over R, solve the equation s*t + r*sin(t) = 0 for r and s for all t that is not the trivial solution (r,s)=(0,0). \nIn other words, if f(t)=sin(t) and g(t)=t were linearly dependent, then there is some non-zero slope s, s*g(t) = s*t and some non-zero amplitude r, r*f(t) = r*sin(t) such that the graphs of these functions completely coincide for all t. Since f is non-linear and g is, this is nonsense and the functions are linearly independent.
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
938958
No, it would have to be the square of something like ax+b, whose square is a^2x+2abx+b^2.  You would have to have a=1 or -1 and b=i or -i, but then 2ab would be imaginary, and not -2.
857762
Not sure if it's the shortest or nicest way, but I was able to get to a solution using substitution u=x^2, and v=y^2, then adding 1/4 to the v side to make it a quadratic equation, solving for v, then re-substituting, and solving for y.\n\nI got y = sqrt ( sqrt (x^2-x^4+1/4) + 1/2 )\n\nYou should be able to plot that ...
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
4959
=8(y^3)z + 12(y^2)(z^2) - 20y^2