If $α,β,γ,δ$ be the roots of the equation $x^4 + px^3 + qx^2 + rx + s = 0$, then find in terms of $p,q,r,s$ the value of $\sum α^4$

Question

If $α,β,γ,δ$ be the roots of the equation $x^4 + px^3 + qx^2 + rx + s = 0$ ,then find in terms of $p,q,r,s$ the value of $\sum \alpha^4$

My general strategy was transforming the equation to one whose roots are $\alpha^4,etc$ but it seems to be impossible as the transformed equation with $y=\alpha^4(\implies y^{1/4}=\alpha)$ is $y+py^{3/4}+qy^{1/2}+ry^{1/4}+s=0,$ which is not even a polynomial in $y$ and hence we can't apply Vieta's formulas.

Next, I tried simplifying the expression $\sum \alpha^4$ but, it doesn't turn out favourable as a huge calculation appears which couldn't be simplified more and no desirable form was obtained and this was in vain too. I don't understand how to approach it...

Answer 1

HINT…use $$\Sigma\alpha^4=-p\Sigma\alpha^3-q\Sigma\alpha^2-r\Sigma\alpha-s\Sigma 1$$ And $$\Sigma\alpha^3=-p\Sigma\alpha^2-q\Sigma\alpha-r\Sigma 1-s\Sigma\frac{1}{\alpha}$$ And all the remaining terms can easily be obtained using Vieta etc.

Answer 2

We have $$x^4+qx^2+s=-x(px^2+r)\implies(x^4+qx^2+s)^2=(-x)^2(px^2+r)^2$$

Replacing $x^2$ with $y$

$$(y^2+qy+s)^2=y(py+r)^2\iff y^4+(2q-p^2)y^3+y^2(q^2+2s-2pr)+y(2qs-r^2)+s^2=0$$ whose roots are $\alpha^2,\beta^2,\gamma^2,\delta^2$

$$\implies\alpha^4+\beta^4+\gamma^4+\delta^4=(p^2-2q)^2-2(q^2+2s-2pr)=?$$

Answer 3

From Vieta's formulas, we have:

$$p = - (\alpha + \beta + \gamma + \delta) \tag{1}$$ $$q = (\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) \tag{2}$$ $$r = - (\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta) \tag{3}$$ $$s = \alpha\beta\gamma\delta \tag{4}$$

Squaring equation (1) gives:

$$p^2 = (\alpha + \beta + \gamma + \delta)^2$$ $$p^2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)$$

But using equation (2), this can be simplified to:

$$p^2 - 2q = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \tag{5}$$

Now let's try cubing equation 1.

$$-p^3 = (\alpha + \beta + \gamma + \delta)^3$$ $$-p^3 = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 + 3\alpha^2(\beta + \gamma + \delta) + 3\beta^2(\alpha + \gamma + \delta) + 3\gamma^2(\alpha + \beta + \gamma) + 3\delta^2(\alpha + \beta + \gamma) + 6(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta)$$ $$-p^3 = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 + 3\alpha^2(-p - \alpha) + 3\beta^2(-p - \beta) + 3\gamma^2(-p - \gamma) + 3\delta^2(-p - \delta) - 6r$$ $$-p^3 = -2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) - 3p(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) - 6r$$ $$-p^3 = -2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) - 3p(p^2-2q) - 6r$$ $$-p^3 + 3pq - 3r = \alpha^3 + \beta^3 + \gamma^3 + \delta^3 \tag{6}$$

Finally, raise equation (1) to the fourth power.

$$p^4 = (\alpha + \beta + \gamma + \delta)^4$$ $$p^4 = \alpha^4 + \beta^4 + \gamma^4 + \delta^4 + 4\alpha^3(\beta + \gamma + \delta) + 4\beta^3(\alpha + \gamma + \delta) + 4\gamma^3(\alpha + \beta + \delta) + 4\delta^3(\alpha + \beta + \gamma) + 6(\alpha^2\beta^2 + \alpha^2\gamma^2 + \alpha^2\delta^2 + \beta^2\gamma^2 + \beta^2\delta^2 + \gamma^2\delta^2) + 12\alpha^2(\beta\gamma + \beta\delta + \gamma\delta) + 12\beta^2(\alpha\gamma + \alpha\delta + \gamma\delta) + 12\gamma^2(\alpha\beta + \alpha\delta + \beta\delta) + 12\delta^2(\alpha\beta + \alpha\gamma + \beta\gamma) + 24\alpha\beta\gamma\delta$$ $$p^4 = \alpha^4 + \beta^4 + \gamma^4 + \delta^4 + 4\alpha^3(-p - \alpha) + 4\beta^3(-p - \beta) + 4\gamma^3(-p - \gamma) + 4\delta^3(-p - \delta) + 3(\alpha^2(p^2 - 2q - \alpha^2) + \beta^2(p^2 - 2q - \beta^2) + \gamma^2(p^2 - 2q - \gamma^2) + \delta^2(p^2 - 2q - \delta^2)) + 12\alpha^2(-r - \alpha(-p-\alpha)) + 12\beta^2(-r - \beta(-p - \beta)) + 12\gamma^2(-r - \gamma(-p - \gamma)) + 12\delta^2(-r-\delta(-p - \delta)) + 24s$$ $$p^4 = \alpha^4 + \beta^4 + \gamma^4 + \delta^4 - 4\alpha^3p - 4\alpha^4 - 4\beta^3p - 4\beta^4 - 4\gamma^3p - 4\gamma^4 - 4\delta^3p - 4\delta^4 + 3\alpha^2p^2 - 6\alpha^2q - 3\alpha^4 + 3\beta^2p^2 - 6\beta^2q - 3\beta^4 + 3\gamma^2p^2 - 6\gamma^2q - 3\gamma^4 + 3\delta^2p^2 - 6\delta^2q - 3\delta^4 - 12\alpha^2r + 12\alpha^3 p + 12\alpha^4 - 12\beta^2r + 12\beta^3 p + 12\beta^4 - 12\gamma^2r + 12\gamma^3 p + 12\gamma^4 - 12\delta^2r + 12\delta^3 p + 12\delta^4 + 24s$$ $$p^4 = 6(\alpha^4 + \beta^4 + \gamma^4 + \delta^4) + 8p(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) + (3p^2 - 6q - 12r)(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) + 24s$$ $$p^4 = 6(\alpha^4 + \beta^4 + \gamma^4 + \delta^4) + 8p(-p^3 + 3pq - 3r) + (3p^2 - 6q - 12r)(p^2 - 2q) + 24s$$ $$p^4 = 6(\alpha^4 + \beta^4 + \gamma^4 + \delta^4) - 8p^4 + 24p^2q - 24pr + 3p^4 - 6p^2q - 6p^2q + 12q^2 - 12p^2r + 24qr + 24s$$ $$\boxed{\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = p^4 - 2p^2q + 2p^2r - 2q^2 + 4pr - 4qr - 4s}$$

Answer 4

There is a direct formula (see (1) below) issued from Newton identities that can be found here.

In order to understand it, we must give the notations used in this article.

• Coefficients $e_k$ are equal to $(-1)^k$ times the coefficients of the initial polynomial, namely :

$$p(x)=x^4-e_1x^3+e_2x^2-e_3x+e_4$$

• Notation $p_k$ is for the sum of the $k$-th powers of the roots. Then :

$$p_{4}=e_{1}p_{3}-e_{2}p_{2}+e_{3}p_{1}-4e_{4}=e_{1}^{4}-4e_{1}^{2}e_{2}+4e_{1}e_{3}+2e_{2}^{2}-4e_{4}\tag{1}$$

giving the final result :

$$p_4=p^4-4p^2q+4pr+2q^2-4s$$

in coincidence with the result of @Lab Bhattacharjee.