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 =0 (a)^ 3 (b)^ 3 (c)^=3abc
Without finding the cubes factorise (x-2 y)^3+(2y-z)^3+(z-x)^3- Without finding cubes, factorise 1(xy)3 (yz)3 (zx)3 2(x2y)3 (2y3z)3 (3zx)3 Get the answers you need, now! (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) That is it no constraints etc It mentions "This can be done by expanding out the brackets, but there is a more elegant solution" Homework Equations The Attempt at a Solution First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agree
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Factor Consider x^ {3}y^ {3}z^ {3}3xyz as a polynomial over variable x Consider x 3 y 3 − z 3 3 x y z 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^ {3} and m divides the constant factor y^ {3}z^ {3} One such factor is xyzSuppose we wish to find an infinite set of solutions of the equation x^3 x y^3 y = z^3 z (1)where x, y, z are integers greater than 1 If z and x are both odd or both even, we can define integers u and v such that z=uv and x=uv Substituting into equation (1) gives y^3 y = 2v(3u^2 v^2 1)Since v divides the righthand side, it would be nice if it alsodivides the lefthand side,
Incoming Term: (x-y)^3+(y-z)^3+(z-x)^3 factorise, without finding the cubes factorise (x-y)^3+(y-z)^3+(z-x)^3, without finding the cubes factorise (x-2 y)^3+(2y-z)^3+(z-x)^3, factorise (2x-y-z)^3+(2y-z-x)^3+(2z-x-y)^3,
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