How I can find the value of $abc$ using the given equations? If I have been given the value of 
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of 
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried by squaring the equations.
But could not get it.
Thanks in advance.
 A: Isaac Newton could help you with this. If
$$
\eqalign{
e_0 &= 1 \\
e_1 &= a+b+c \\
e_2 &= ab+bc+ca \\
e_3 &= abc
}
\qquad
\eqalign{
p_1 &= a+b+c \\
p_2 &= a^2+b^2+c^2 \\
p_3 &= a^3+b^3+c^3
}
$$
then he showed that
$$
\eqalign{
     e_1 &= p_1 \\
2 \, e_2 &= e_1p_1-p_2 \\
3 \, e_3 &= e_2p_1-e_1p_2+p_3 \\
}
$$
(which solves your problem) and (incidently)
$$
\eqalign{
 p_1 &= p_1 \\
 p_2 &= e_1p_1-2 \, e_2 \\
 p_3 &= e_1p_2-e_2p_1+3 \, p_3. \\
}
$$
Spoiler below...

 $$\eqalign{e_1 &= p_1 &= 1 \\2 \, e_2 &= e_1p_1-p_2 &= 1\cdot1-9 =-8  &\implies e_2=-4 \\3 \, e_3 &= e_2p_1-e_1p_2+p_3 &=-4\cdot1-1\cdot9+1=-12 &\implies e_3=-4 \\}$$

The formulas Newton found are called Newton's identities,
or the Newton–Girard formulae, and relates two kinds of symmetric polynomials: (1) the (homogeneous) sums of $k^\text{th}$ powers, $p_k$,
of some number of indeterminates $a,b,c\dots$
and (2) the sums of products of each $k$ indeterminates,
which are denoted by $e_k$ and are called elementary symmetric polynomials.
It's a very handy trick, and generalizes to
$n$ indeterminates and $1\le k\le n$, but
it is not usually covered in precalculus.
A: You can get a term involving $abc$ by cubing $a+b+c$:
$$\begin{align*}
(a+b+c)^3 &= (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3\\
&= a^3+3a^2b+3ab^2 + b^3 + 3a^2c+\color{blue}{6abc} + 3b^2c + 3ac^2 + 3bc^2 + c^3.
\end{align*}$$
Now use the other information you have to try to find the value of $abc$.
For example, you know this whole thing equals $(a+b+c)^3 = 1$. You also know the value of $a^3+b^3+c^3$...
A: In general, $$a^n + b^n + c^n = \sum_{i+2j+3k=n} (-1)^j \frac{n}{i+j+k}{i+j+k\choose i,j,k}s_1^is_2^js_3^k$$
where the sum is over non-negative $i,j,k$, and where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$.
In particular, when $n=3$ there are only three triples $(i,j,k)=(3,0,0),(1,1,0),(0,0,1)$, and you get:
$$a^3+b^3+c^3 = (a+b+c)^3 - 3(ab+ac+bc)(a+b+c) + 3abc$$
Now solve for $abc$.
A: If $a,b,c$ solve the equation $x^3+mx^2+nx+p=0$ then you know that $S_3+mS_2+nS_1+3p=0$, where $S_i=a^i+b^i+c^i$. From the sum you find who $m$ is. The expression of $n$ is $ab+bc+ca$. You can find $p$ substituting all the values in the equation. Then you can find the product, which is $-p$ and eventually solve the equation.

Alternatively, you can find the product using the formula
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) $$
A: First $\displaystyle{(a+b+c)^2 = (a^2+b^2+c^2)+2(ab+bc+ca)}$, which implies $1 = 9+2X$ where $X=(ab+bc+ca) \implies (ab+bc+ca)=-4$
Using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, and substituting the known values  $1-3Y = 9+4$ and solve for $Y=abc$
Note: You can always check with Wolfram Alpha if your answer is correct (not to solve your problem) Check http://tinyurl.com/7n6ey2t
