You are not completely clear on your question. However,
a^(2) + b - c /m is one answer ( NB The 'm' is just divided into the 'c').
Another answer is
[ a^(2) + b - c ] / m This means that the 'm' is divided into the sum of all the terms in the (square) brackets.
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