Sum of powers

$a+b+c=1$
$a^2+b^2+c^2=21$
$a^3+b^3+c^3=55$

$(a+b+c)^2-2(ab+bc+ca)=21$
$=\Rightarrow (ab+bc+ca)=-10$

$a^3+b^3+c^3-3abc=55-3abc$
$=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))=55-3abc$
$=\Rightarrow (abc)=8$

$(a-1)(b-1)(c-1)$
$=(abc)-(ab+bc+ca)+(a+b+c)-1$
$=\Rightarrow 8+10+1-1=\boxed{18}$

Let $a,b,c$ be the roots of a cubic polynomial $p(x)$ .

Now, by Newton's Sums, we get one of those polynomials to be equal to, $p(x)=(x^3-x^2-10x-8)$ .

It would be of the form, $p(x)=-(a-x)(b-x)(c-x)$ ,for some constant $k$ .

So we get $p(1)=-(a-1)(b-1)(c-1)=\boxed{-18}$

Another solution is using the identity $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc.$ From given conditions, we have $55=1-3(ab+bc+ca)+3abc.$ So we get that $abc-(ab+bc+ca)=18$ . The desired expression is $abc-(ab+bc+ca)+(a+b+c)-1=18+1-1=18.$ Note that this solution doesn't need $a^2+b^2+c^2=21$ .