Cubic Equation Challenge: What is the value of $mn^2+nk^2+km^2$?

[anemone]

For all real $m,\,n,\,k$ where $m>n>k$, these three real numbers are the roots for the equation $x^3-2x^2-x+1=0$.

Evaluate $mn^2+nk^2+km^2$.

 

[Opalg]

If $m,\,n,\,k$ (all real), where $m>n>k$, are the roots for the equation $x^3-2x^2-x+1=0$, evaluate $mn^2+nk^2+km^2$.

I have slightly amended the wording of the problem to avoid the impression that it is saying that something is true "for all real $m$, $n$ and $k$".
[sp]Outline proof (I don't have time to write it out in full): Let $f(x) = x^3-2x^2-x+1$. Then $f(-1) = -1$, $f(0) = 1$, $f(1) = f(2) = -1$, $f(3) = 7$. It follows that the roots must satisfy $-1<k<0<n<1$ and $2<m<3$. This implies that $mn^2+nk^2+km^2 < 0$.

Next, Let $S_1 = mn^2+nk^2+km^2$ and $S_2 = m^2n+n^2k+k^2m$. Then $(nk+km+mn)(m+n+k) = S_1 + S_2 + 3mnk.$ But $m+n+k = 2$, $nk+km+mn = -1$ and $mnk = -1$ (symmetric functions of the roots). Therefore $S_1+S_2 = -2+3=1.$

The product $S_1S_2$ is $$(mn^2+nk^2+km^2)(m^2n+n^2k+k^2m) = (n^3k^3 + k^3m^3 + m^3n^3) + 3m^2n^2k^2 + mnk(m^3+n^3+k^3).$$ To evaluate that, notice that the equation with roots $m^3$, $n^3$, $k^3$ is given by letting $y=x^3$ in the original equation, which then becomes $y +1 = 2y^{2 /3} + y^{1 /3}$. Cube both sides to see that this gives $y^3 - 11y^2 - 4y + 1 = 0$. Therefore $m^3+n^3+k^3 = 11$ and $n^3k^3 + k^3m^3 + m^3n^3 = -4.$ Substitute those values into the above displayed equation to get $S_1S_2 = -4+3-11 = -12$.

Thus the equation with roots $S_1$ and $S_2$ is $\lambda ^2 - \lambda - 12 = 0$, with roots $\lambda=4$ and $\lambda = -3$. But we know that $S_1<0$, so the answer has to be that $S_1 = -3.$[/sp]

 

[anemone]

I have slightly amended the wording of the problem to avoid the impression that it is saying that something is true "for all real $m$, $n$ and $k$".

Thanks, Opalg for amending the wording of the problem to make it sound for me. I appreciate that!

[sp]Outline proof (I don't have time to write it out in full): Let $f(x) = x^3-2x^2-x+1$. Then $f(-1) = -1$, $f(0) = 1$, $f(1) = f(2) = -1$, $f(3) = 7$. It follows that the roots must satisfy $-1<k<0<n<1$ and $2<m<3$. This implies that $mn^2+nk^2+km^2 < 0$.

Next, Let $S_1 = mn^2+nk^2+km^2$ and $S_2 = m^2n+n^2k+k^2m$. Then $(nk+km+mn)(m+n+k) = S_1 + S_2 + 3mnk.$ But $m+n+k = 2$, $nk+km+mn = -1$ and $mnk = -1$ (symmetric functions of the roots). Therefore $S_1+S_2 = -2+3=1.$

The product $S_1S_2$ is $$(mn^2+nk^2+km^2)(m^2n+n^2k+k^2m) = (n^3k^3 + k^3m^3 + m^3n^3) + 3m^2n^2k^2 + mnk(m^3+n^3+k^3).$$ To evaluate that, notice that the equation with roots $m^3$, $n^3$, $k^3$ is given by letting $y=x^3$ in the original equation, which then becomes $y +1 = 2y^{2 /3} + y^{1 /3}$. Cube both sides to see that this gives $y^3 - 11y^2 - 4y + 1 = 0$. Therefore $m^3+n^3+k^3 = 11$ and $n^3k^3 + k^3m^3 + m^3n^3 = -4.$ Substitute those values into the above displayed equation to get $S_1S_2 = -4+3-11 = -12$.

Thus the equation with roots $S_1$ and $S_2$ is $\lambda ^2 - \lambda - 12 = 0$, with roots $\lambda=4$ and $\lambda = -3$. But we know that $S_1<0$, so the answer has to be that $S_1 = -3.$[/sp]

Well done, Opalg! Since the solutions that I have are more tedious than yours, hence I don't think I will show them here.:)