Because there is no way to define the divisors, the equations cannot be evaluated.
A2 + B2 = C2 If C=8, then A2 + B2 = 64
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
l a2 b2 is c2!!Its completely norma
a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z
It is a*a + b*b + c*c
(a+b+c) (a+b-c)
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
Because there is no way to define the divisors, the equations cannot be evaluated.
int x = 2 * (a + b);
A2 + B2 = C2 If C=8, then A2 + B2 = 64
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
l a2 b2 is c2!!Its completely norma
a2 + b2 + c2 - ab - bc - ca = 0 => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0 Rearranging, a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 0 => (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0 or (a - b)2 + (b - c)2 + (c - a)2 = 0 so a - b = 0, b - c = 0 and c - a = 0 (since each square is >=0) that is, a = b = c
a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z
{ int a,b; { a=a^2; } { b=b^2; } { c=a^2+b^2+2*a*b; print f("%d%d%d",&c); get ch(); } ]
5 - a - 1 + c - 6a2 + a2