If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$

Question

Question:

If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma}-{\alpha})^2$

Now, the normal way to solve this question would be to use the theory of equations and find the sum of roots taken one at a time, two at a time and three at a time. Using this approach, we get the answer as $(x+1)^3+3(x+1)^2+27=0$. However, I feel that this is a very lengthy approach to this problem. Is there an easier way of doing it?

Answer 1

Let $a,b,c$ be the roots of $x^3+x+1=0$ so we have $a+b+c=0, ab+bc+ca=1,abc=-1$, so $a^2+b^2+c^2=-2$ and $c^3=-c-1$

We would explore a transformation from $x$ to $y$ to get the required cubic equation of $y$. Let $$y=(a-b)^2=a^2+b^2-2ab\implies y=-2-c^2+2/c \implies c=\frac{3}{1+y}$$ Replacing $c$ by $x$ we get the required transformation $x=\frac{3}{1+y}$, putting it in the given $x$ equation, we get: $$\frac{27}{(1+y)^3}+\frac{3}{(1+y)}+1=0 \implies y^3+6y^2+9y+31=0,$$ which is the required cubic equation.

Answer 2

Final constant term $1+3+27=31$ can be obtained at once (or checked at once) by considering that it is the opposite of the product of roots

$$(({\alpha}-{\beta})({\beta}-{\gamma})({\gamma}-{\alpha}))^2$$

which is the classical discriminant $-(4p^3+27q^2)$ of a reduced 3rd degree equation $X^3+pX+q=0$ with $p=q=1$. (https://en.wikipedia.org/wiki/Discriminant#Degree_3)

Answer 3

Hint:

Let $y=(a-b)^2=(a+b)^2-4ab=(-c)^2-\dfrac4{-c}$ as $abc=-1, a+b=-c$

$$\iff c^3-cy+4=0\ \ \ \ (1) $$

Again we have $$c^3+c+1=0\ \ \ \ (0)$$

Solve the two simultaneous equations for $c,c^3$ and use $c^3=(c)^3$ to eliminate $c$

Answer 4

The standard method: $$a+b+c=0; ab+bc+ca=1; abc=-1;\\ a^2+b^2+c^2=-2;a^2b^2+b^2c^2+c^2a^2=1;a^4+b^4+c^4=2;\\ a^3=-a-1.$$ First coefficient: $$(a-b)^2+(b-c)^2+(c-a)^2=\\2(a^2+b^2+c^2)-2(ab+bc+ca)=-6$$ Second coefficient: $$(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=\\ a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)-\\ 2(a^3(\underbrace{b+c}_{-a})+b^3(\underbrace{c+a}_{-b})+c^3(\underbrace{a+b}_{-c}))=\\ 3(a^2b^2+b^2c^2+c^2a^2+a^4+b^4+c^4)=9$$ Third coefficient: $$(a-b)^2(b-c)^2(c-a)^2=\\ \small{(a^2+b^2-2ab)(b^2+c^2-2bc)(c^2+a^2-2ac)=\\ (c^2-4ab)(a^2-4bc)(b^2-4ac)=\\ 16 a^4 b c - 4 a^3 b^3 - 4 a^3 c^3 - 63 a^2 b^2 c^2 + 16 a b^4 c + 16 a b c^4 - 4 b^3 c^3=\\ -16(a^3+b^3+c^3)-4(a^3b^3+b^3c^3+c^3a^3)-63=\\ -16(-a-b-c-3)-4((-a-1)(-b-1)+\\ (-b-1)(-c-1)+(-c-1)(-a-1))-63=}\\ 48-4(1+3)-63=-31.$$ Hence, the equation is: $x^3+6x^2+9x+31=0$.