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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.

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You can also cube your first equation, and using $ab+bc+ac$ you found, you should be able to find $abc$. – Lazar Ljubenović Mar 15 '12 at 16:17
Here is a solution I like, but way overkill to write as an answer. Let $A$ be any $3x3$ matrix which has eigenvalues $a,b,c$. Then you are given $tr(A), tr(A^2), tr(A^3)$ and need to find $\det A$. Write the characteristic equation of $A$ and apply trace on it.... – N. S. Mar 15 '12 at 17:09
I think you have got enough answers,you should go ahead and chose whichever is your acceptable answer – Jeremy Carlos Mar 15 '12 at 17:11
up vote 6 down vote accepted

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 $a^3+b^3+c^3$...

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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.

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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 $abc$.

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

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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

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