Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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
1  
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
add comment

5 Answers

up vote 5 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 $a^3+b^3+c^3$...

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $abc$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.