The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$ 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?
 A: 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 $r$.
A: $la+mbc,lb+mca,lc+mab$ 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
$x^3 - (l(a+b+c) + m(bc+ca+ab)))x^2 + ((la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)) x - (la+mbc)(lb+mca)(lc+mba)$
