Suppose a,b,c∈Isuch that the greatest common divisor of x2+ax+b and x2+bx+c is (x+1)and the least common multiple of x2+ax+b and x2+bx+c is x3−4x2+x+6.Then the value of 10 abc is equal to ______.

x2+ax+b=(x+1)(x+b)⇒b+1=a    (1)Also,  x2+bx+c=(x+1)(x+c)⇒c+1=bor            b+1=c+2                             (2)Hence   b+1=a=c+2Also   (x+1)(x+b)(x+c)=x3−4x2+x+6⇒    x3+(1+b+c)x2+(b+bc+c)x+bc≡x3−4x2+x+6⇒    1+b+c=−4⇒    2c+2=−4⇒c=−3;b=−2 and a=−1    10abc=−60