Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
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Is there no constant term? (There's a stray plus sign in there, if not.) Also, why is this in (differential-equations)? – Brian Tung Aug 06 '15 at 21:42
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@BrianTung my mistake:), the question is correct, and I also got a solution for it :) – Oiue Aug 06 '15 at 21:43
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2$4x$ plus what? – Kaster Aug 06 '15 at 21:45
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But plus what at the end? – Oussama Boussif Aug 06 '15 at 21:45
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Sorry guys, typo has been fixed – Oiue Aug 06 '15 at 21:47
3 Answers
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If $a,b,c$ are the roots of $p(x)=8x^3-4x^2-4x+1$ then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the roots of $$ q(x) = x^3-4x^2-4x+8 $$ so: $$ \frac{1}{a^3} = \frac{4}{a^2}+\frac{4}{a}- 8 $$ and: $$\sum_{cyc}\frac{1}{a^3}=4\sum_{cyc}\frac{1}{a^2}+4\sum_{cyc}\frac{1}{a}-24$$ then Viète's theorem applied to $q(x)$ gives: $$\sum_{cyc}\frac{1}{a^3}=4(4^2-2\cdot(-4))+4(4)-24 = \color{red}{88}.$$
With a bit of experience, one may recognize $p(x)$ as the minimal polynomial of $\alpha=-\cos\frac{2\pi}{7}$, whose conjugates are $-\cos\frac{4\pi}{7}$ and $-\cos\frac{6\pi}{7}$. By this way, the problem is equivalent to proving a not-so-difficult trigonometric identity.
Another neat trick is the following: given $p(x)=\left(1-\frac{x}{a}\right)\left(1-\frac{x}{b}\right)\left(1-\frac{x}{c}\right)$, $$ \log p(x) = \sum_{cyc}\log\left(1-\frac{x}{a}\right) = -\sum_{cyc}\left(\frac{x}{a}+\frac{x^2}{2a^2}+\frac{x^3}{3a^3}+\ldots\right) $$ hence $\sum_{cyc}\frac{1}{a^3}$ is minus three times the coefficient of $x^3$ in the Taylor series of $\log p(x)$ in a neighbourhood of $x=0$, or: $$ \sum_{cyc}\frac{1}{a^3} = -\frac{1}{2}\left.\frac{d^3}{dx^3}\log p(x)\right|_{x=0}, $$ and we just need to evaluate: $$\frac{\left(4+8 x-24 x^2\right)^3}{\left(1-4 x-4 x^2+8 x^3\right)^3}+\frac{3 (-4+24 x) \left(-4-8 x+24 x^2\right)}{\left(1-4 x-4 x^2+8 x^3\right)^2}-\frac{24}{1-4 x-4 x^2+8 x^3} $$ at $x=0$ to get $4^3+3\cdot 4^2-24=\color{red}{88}$ just as before.
Jack D'Aurizio
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you always give as a short answer, genius man, you don't want to see my solution it's too long, (+1) – Oiue Aug 06 '15 at 21:56
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1@man_in_green_shirt: $\sum_{cyc}\frac{1}{a}$ is a shorthand notation for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. In general, $\sum_{cyc}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b)$. – Jack D'Aurizio Aug 06 '15 at 22:12
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1@joriki: we need the sum of the squares of the roots of a polynomial. We write $u^2+v^2+w^2$ as $(u+v+w)^2-2(uw+uv+vw)$ and recall that we can read both $(u+v+w)$ and $(uv+vw+uw)$ in the coefficients of the polynomial. – Jack D'Aurizio Aug 06 '15 at 22:13
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A systematic way to compute sums of the form $a^{-k} + b^{-k} + c^{-k}$ is using Newton's identities.
Notice $$\begin{align} a, b, c \text{ roots of } p(x) &= 8x^3 - 4x^2 - 4x + 1\\ \implies\quad \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ roots of } q(x) &= x^3 p(\frac{1}{x}) = x^3 - \color{red}{4} x^2 - \color{green}{4}x + \color{blue}{8} \end{align} $$
Let $p_k = a^{-k} + b^{-k} + c^{-k}$ for $k = 1,2,3$. Newton's identities tell us:
$$\begin{cases} p_1 - \color{red}{4} \times 1 &= 0\\ p_2 - \color{red}{4} p_1 - \color{green}{4}\times 2 &= 0\\ p_3 - \color{red}{4} p_2 - \color{green}{4} p_1 + \color{blue}{8}\times 3 &= 0 \end{cases} \quad\implies\quad \begin{cases} p_1 &= 4\\ &\;\Downarrow\\ p_2 &= 4\times \underbrace{4}_{p_1} + 4\times 2 = 24\\ &\;\Downarrow\\ p_3 &= 4\times \underbrace{24}_{p_2} + 4\times \underbrace{4}_{p_1} - 8\times 3 = 88 \end{cases} $$
achille hui
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It has been observed by others that the reciprocals $x_i$ $(1\leq i\leq 3)$ of $a$, $b$, $c$ are the solutions of the equation $$x^3-4x^2-4x+8=0\ .\tag{1}$$ The power sum $$p_3:=x_1^3+x_2^3+x_3^3={1\over a^3}+{1\over b^3}+{1\over c^3}$$ is a symmetric function of the $x_i$. Therefore it can be expressed as a polynomial in terms of the elementary functions $\sigma_j$ $(1\leq j\leq 3)$ of the $x_i\,$: One easily verifies that $$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3\ ,$$ or $$p_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\ .$$ Now $(1)$ and Vieta's theorem tell us that $$\sigma_1=4,\quad \sigma_2=-4,\qquad \sigma_3=-8\ .$$ It follows that $$p_3=64-3\cdot4\cdot(-4)+3\cdot(-8)=88\ .$$
Christian Blatter
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