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If $a,b,c$ are three distinct real numbers for which there exists $x,y,z$ such that

$x+y+z=1$

$ax+by+cz=t$

$a^2x+b^2y+c^2z=t^2$and $a^3x+b^3y+c^3z-t^3=p(t),$ polynomial in $t$

then number of solution of $p(t) =0$

i want be able to go ahead, could some help me with this, thanks

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    You can solve the first 3 equations in x,y,z using matrix algebra, giving you a general solution in terms of a,b,c,t. – 2017-01-10

2 Answers 2

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Solving the first 3 equations in $x,y,z$ is equivalent to solving: $$ \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \\ \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ t \\ t^2 \\ \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} $$ This will give you solutions for $x,y,z$ in terms of $a,b,c,t$ after you calculate the inverse on the left:

$x = \frac {bc - tb - tc + t^2}{a^2 - ab - ac + bc}$ $y = \frac {ta + tc - ac - t^2}{ab - ac -b^2 + bc}$ $z = \frac {ab - ta - tb + t^2}{ab - ac - bc + c^2}$

This is the unique solution, and by substituting these into the fourth equation you get: $p(t) = \frac {a^3(bc - tb - tc + t^2)}{a^2 - ab - ac + bc} + \frac {b^3(ta + tc - ac - t^2)}{ab - ac -b^2 + bc} + \frac {c^3(ab - ta - tb + t^2)}{ab - ac - bc + c^2} - t^3$

This is factorizable as $p(t) = \frac {a^3(t-b)(t-c)(a-b)(b-c)(c-a)(c-b) + b^3(t-a)(c-t)(a-b)(a-c)(c-a)(c-b) + c^3(t-a)(t-b)(a-b)(b-c)(a-b)(a-c)}{(a-b)(a-c)(a-b)(b-c)(c-a)(c-b)} - t^3$

with solutions $t = a,b,c$ which are the only solution as it is a cubic.

sorry for the staggered solution, I was moving around.

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    I came to the same answer (three solutions of t) with algebra and simultaneous equations, but wasn't sure how it answered the question of "number of solution of p(t)=0" which I thought meant that the polynomial had no solutions. – 2017-01-10
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    p(t) on its own is just a polynomial. It must be asking for solutions of "p(t) = 0". – 2017-01-11
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We are given $P(t)$ in this problem and need to prove that it has no solutions.

We might do this by inspecting the behaviour of the function $P(t)$, either graphically or using calculus/other techniques.

Hint

$P(t)$ here contains terms in $x, y, z$, various combinations of which form a power of $t$.