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

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Challenge Cubic
AI Thread Summary
The cubic equation \(x^3 - 2x^2 - x + 1 = 0\) has roots \(m, n, k\) where \(m > n > k\) and these roots fall within specific intervals. Evaluating the expression \(mn^2 + nk^2 + km^2\) leads to the conclusion that it is less than zero. By defining \(S_1 = mn^2 + nk^2 + km^2\) and \(S_2 = m^2n + n^2k + k^2m\), a relationship between these sums is established, ultimately leading to the equation \(\lambda^2 - \lambda - 12 = 0\). The roots of this equation are \(4\) and \(-3\), and since \(S_1\) must be negative, the final result is \(S_1 = -3\).
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $mn^2+nk^2+km^2$.
 
Mathematics news on Phys.org
anemone said:
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]
 
Opalg said:
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!:o

Opalg said:
[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.:)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top