设函数f(x)=(x-a)(x-b)(x-c),a,b,c属于R,

a = 0，f(x) = x( x - b )( x - c ) = (1/2)[ 2x( b - x )( c - x ) ]；因为2x +( b - x ) + ( c - x )是常数，所以2x = b - x = c - x 时，2x( b - x )( c - x ) 有最大值；即 b = c，2x = b - x，x = b/3 时，f(x) 有最大值。故 M =(b/3) * ( 2b/3 ) * ( 2b/3 ) = 4b^3/27≤ 4/27 。