b+b+b+c+c+c+c =3b+4c
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
(a + b+ c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3b2c + 3bc2 + 3c2a + 3ca2 + 6abc
2b + 2c or 2(b + c)
And how does this relate to coins?
(a+b+c)²=a²+b²+c²+ 2ab+2bc+2ac
b+b+b+c+c+c+c =3b+4c
b + b + b + c + c + c + c = 3b + 4c
(2a + c)(4a2 - 2ac + c2)
(a+b-c)2 = a2 + b2 +c2 +2ab - 2bc - 2ac
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
(a + b+ c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3b2c + 3bc2 + 3c2a + 3ca2 + 6abc
It is impossible to give any decimal/numeric value if we are not given the values of at least one variable, so the answer is B + B + B + C + C + C.
2b + 2c or 2(b + c)
(a+b+c) 2=a2+ab+ac+ba+b2+bc+ca+cb+c2a2+b2+c2+2ab+2bc+2ca [ ANSWER!]
(c + 3)(c^2 - 3c + 9)
And how does this relate to coins?