Find the value of $\sum \frac{1}{\beta+\gamma}$.

Question

If $\alpha,\beta,\gamma$ are the roots of the cubic equation $px^3+3qx^2+3rx+s=0$ then find the value of $\sum \dfrac{1}{\beta+\gamma}$.

TRY:

We have $\sum \alpha=\dfrac{-3q}{p},\sum \alpha\beta =\dfrac{3r}{p},\alpha\beta\gamma=\dfrac{s}{p}.$

Now $\sum \dfrac{1}{\beta+\gamma}=\dfrac{1}{\alpha+\beta}+\dfrac{1}{\beta+\gamma}+\dfrac{1}{\gamma+\alpha}$.

But how to proceed from here? Please help.

Answer 1

If $f(x) = px^3 + 3qx^2 + 3rx + s = p(x-\alpha)(x-\beta)(x-\gamma)$, then $\alpha + \beta + \gamma = -\frac{3q}{p}$. Furthermore, we have the identity

$$\frac{f'(x)}{f(x)} = \sum_{cyc} \frac{1}{x - \alpha}$$ This implies

$$\require{cancel} \sum_{cyc}\frac{1}{\beta+\gamma} = \sum_{cyc}\frac{1}{-\frac{3q}{p}-\alpha} = \frac{f'(-\frac{3q}{p})}{f(-\frac{3q}{p})} = \frac{ \color{red}{\cancelto{\frac{9q^2}{p}}{ \color{gray}{ 3p \left(-\frac{3q}{p}\right)^2 + 6q \left(-\frac{3q}{p}\right) }}} + 3r}{ \color{red}{\cancelto{0}{\color{gray}{p\left(-\frac{3q}{p}\right)^3 + 3q\left(-\frac{3q}{p}\right)^2}}} + 3r\left(-\frac{3q}{p}\right) + s}$$ Multiply both numerator and denominator by $p$, this can be simplified to $\displaystyle\;\frac{9q^2+3pr}{ps-9qr}\;$.

Answer 2

Hint:

$$y=\dfrac1{\beta+\gamma}=\dfrac1{-\dfrac{3q}p-\alpha}$$

$$\iff\alpha=-\dfrac1y-\dfrac{3q}p=-\dfrac{p+3qy}{yp}$$

Replace this value in the given equation to from a cubic equation in $y$

$$p\left(-\dfrac{p+3qy}{yp}\right)^3+3q\left(-\dfrac{p+3qy}{yp}\right)^2+3r\left(-\dfrac{p+3qy}{yp}\right)+s=0$$

$$\iff-p(p+3qy)^3+3pqy(p+3qy)^2-3p^2ry^2(p+3qy)+sp^3y^3=0$$

$$\iff y^3(-27pq^3+27pq^3-9p^2qr+sp^3)+y^2(-27p^2q^2+18p^2q^2-3p^3r)+\cdots=0$$

Now apply Vieta's formula to find $$\dfrac{1}{\alpha+\beta}+\dfrac{1}{\beta+\gamma}+\dfrac{1}{\gamma+\alpha}=-\dfrac{-3p^2(pr+3q^2)}{p^2(ps-9qr)}=?$$

Answer 3

By the Viete's theorem we have: $$\alpha+\beta+\gamma=-\frac{3q}{p},$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{3r}{p}$$ and $$\alpha\beta\gamma=-\frac{s}{p}.$$ Thus,

$$\sum_{cyc}\frac{1}{\alpha+\beta}=\frac{\sum\limits_{cyc}(\alpha+\beta)(\alpha+\gamma)}{\prod\limits_{cyc}(\alpha+\beta)}=\frac{\sum\limits_{cyc}(\alpha^2+3\alpha\beta)}{(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-\alpha\beta\gamma}=$$ $$=\frac{(\alpha+\beta+\gamma)^2+\alpha\beta+\alpha\gamma+\beta\gamma}{(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-\alpha\beta\gamma}=\frac{\left(-\frac{3q}{p}\right)^2+\frac{3r}{p}}{-\frac{3q}{p}\cdot\frac{3r}{p}+\frac{s}{p}}=\frac{3(pr+3q^2)}{ps-9pr}.$$

Answer 4

Hint:   $\beta+\gamma=-\frac{3q}{p}-\alpha\,$, then calculate $\displaystyle\sum \frac{1}{-\frac{3q}{p}-\alpha}\,$ which works out somewhat easier.

Answer 5

Note as others have done that $\alpha+\beta+\gamma = -\frac {3q}p:=t$ so that $pt=-3q$

Another route through is to find the equation with roots $y=\frac 1{t-x}$ whence $x=t-\frac 1y=\frac {ty-1}y$

So substitute into the original equation and multiply through by $y^3$ $$p(ty-1)^3+3qy(ty-1)^2+3ry^2(ty-1)+sy^3=0$$

Since you want the sum of the roots you only need the coefficients of $y^3$ and $y^2$

For $y^3$ we have:$$pt^3+3qt^2+3rt+s=3rt+s$$(using $pt=-3q$)

For $y^2$ we get:$$-3pt^2-6qt-3r=3qt-3r$$

This approach simplifies the computations somewhat.

Then it is easy to finish.