Please tell how to solve it and approach such type of difficult questions

Question

Please tell how to approach such types of questions with a detailed solution (https://i.stack.imgur.com/O7Lwl.jpg)

If $\alpha ,\beta ,\gamma ,\delta$ are roots of the equation $x^4+ax^2+bx+c=0$

Then find $$\frac{\alpha^7+\beta^7+\gamma^7+\delta^7}{b(c-a^2)}$$

Answer 1

For these type of questions, use the formula :

For a polynomial -

$P(x)=a_0x^n+a_1x^{n-1}+.......a_{n-1}x+a_n$

;

$\sum_{i=0}^{r-1} a_iS_{r-i}+ra_r=0$

Where ;

$S_r=\sum \alpha_i^r$

And, $\alpha_i$ are roots of polynomial $P(x)=0$

Apply this formula for $r=0,1,2,3...m$ till you get value of $S_m$

If you run out of $a_r$ (when $r>n$) multiply the whole polynomial by $x$ , this will introduce on more root of $P(x)=0$ , but the new root would be $x=0$ so it would have no effect on your $S_m$ ( Since $0^m$ i.e. $0$ will be added to $S_m$ which will have no effect)

For this question :

$a_0S_r+a_1S_{r-1}+...ra_r=0$

$r=2$

$a_0S_2+a_1S_1+2a_2=0$

Which gives $S_2=-2a$

$r=3$

$a_0S_3+a_1S_2+a_2S_1+3a_3=0$

Which gives $S_3=-3b$

$r=4$ gives $S_4=2a^2-4c$

$r=5$

Multiply $P(x)$ by $x$

$x^5+ax^3+bx^2+cx=0$

Here, put $r=5$

Note : $a_5=0$ here

$a_0S_5+a_1S_4+a_2S_3+a_3S_2+a_4S_1+5a_5=0$

Which gives $S_5 = 5ab$

Now proceed as I have done above.

This method seems to be lengthy, but believe me it's very helpful and short when you practice for it, specially in those type of questions where numerical values are given.

This method also helps in other problems too, for example

$x+y+z=1$

$x^2+y^2+z^2=2$

$x^3+y^3+z^3=3$

Find : $x^5+y^5+z^5$

Here, assume a cubic equation whose roots are $x,y$ and $z$, and apply the formula I have provided you .This solves the problem in a couple of minutes!

Answer 2

Hint #1:  for each root $\alpha^4 = -a \alpha^2 - b \alpha- c\,$, so:

$$ \begin{align} \alpha^7 & = -a \alpha^5 - b\alpha^4 - c\alpha^3 \\ & = -a(-a\alpha^3-b\alpha^2-c\alpha) - b(-a \alpha^2 - b \alpha- c) - c\alpha^3 \\ & = (a^2-c)\alpha^3 +2ab \alpha^2+(ac+b^2)\alpha+bc \end{align} $$

Hint #2: the quartic has no $x^3$ term so $\alpha+\beta+\gamma+\delta=0\,$. Then, by Newton's relations (or direct calculation):

$$ \begin{alignat}{2} \alpha^2+\beta^2+\gamma^2+\delta^2&=-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta) & = -2a \\ \alpha^3+\beta^3+\gamma^3+\delta^3 &= 3(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta) & = -3b \end{alignat} $$

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[ EDIT ]   Hint #1+2:

$$\require{cancel} \begin{align} \sum \alpha^7 &= (a^2-c)\sum\alpha^3 +2ab \sum\alpha^2+\bcancel{(ac+b^2)\sum\alpha}+4bc \\ &=(a^2-c)\,(-3b)+2ab\,(-2a)+4bc = \;\cdots \end{align} $$