The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$

Question

The roots of the cubic $x^3+qx+r=0$ are $a,b,c$.

How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$?

Can anyone help me to solve this problem?

Answer 1

The long, brute-force way: Expand out $\left(x-(la+mbc)\right)\left(x-(lb+mca)\right)\left(x-(lc+mab)\right)$ to get its full form as a cubic; you should find that all of the coefficients are symmetric functions of $(a,b,c)$. Then use the Fundamental Theorem Of Symmetric Polynomials to express those coefficients in terms of the basic symmetric polynomials $S_1(a,b,c) = a+b+c$, $S_2(a,b,c) = ab+bc+ca$, and $S_3(a,b,c)=abc$. Finally, use the traditional theorems on expressing the coefficients of a polynomial in terms of its roots (e.g., $r=-abc$) to express the coefficients in terms of $q$ and $r$.

Answer 2

$la+mbc,lb+mca,lc+mab$ as our roots means that

From the given equation we have $a+b+c = 0$ since coefficient of $x^2$ is 0. and $ab+ba+bc =q$

$x^3 + px^2+ qx + r$ (independent of the equation in your question)

$p = l(a+b+c) + m(bc+ca+ab)$

$p = l(0) + m(q) = mq$ where q is the coefficient of x in your expression

Similarly (I haven't really expressed it in terms of the coefficients of your given equation),

$q = (la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)$

$r = (la+mbc)(lb+mca)(lc+mba)$

The cubic polynomial with these roots is of the form

$x^3 - (l(a+b+c) + m(bc+ca+ab)))x^2 + ((la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)) x - (la+mbc)(lb+mca)(lc+mba)$