2014/069 Given $P(x)=x^3-6x^2+17x$. If $x=m$, $P(m)=16$ and when $x=n$, $P(n)=20$. Evaluate $m+n$.

we have $P(x) = x^3 - 6x^2 + 17x$

so we have $P(x+2) = x^3 + 5x + 18$ ( I take x+2 to eliminate the $x^2$ term to see in case we get odd function)

now
$P(m) = (m-2)^3 + 5(m-2) + 18 = 16$

or $(m-2)^3 +5 (m-2) = - 2\cdots (1)$

$P(n) = (n-2)^3 + 5(n-2) + 18 = 20$

or $(n-2)^3 + 5(n-2) = 2\cdots (2)$

from (1) and (2) as

$f(x) = x^3 + 5x$

$f(m-2) + f(n-2) = 0$ so $m- 2 + n -2 = 0$

or $m+n = 4$