Evaluate (without a calculator) the product: f(r1+1)⋅f(r2+1)⋅f(r3+1).

[lfdahl]

Given the function:

$$f(x) = x^3+20x-17.$$

Denote the roots of the function: $r_1,r_2$ and $r_3$.

Evaluate (without a calculator) the product: $f(r_1+1) \cdot f(r_2+1) \cdot f(r_3+1)$.

 

[Albert1]

Given the function:

$$f(x) = x^3+20x-17.$$

Denote the roots of the function: $r_1,r_2$ and $r_3$.

Evaluate (without a calculator) the product: $f(r_1+1) \cdot f(r_2+1) \cdot f(r_3+1)$.

my solution:

Spoiler

for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}$
$=\dfrac{27000+52920*20-74088*17}{17-20-1}=\dfrac{-174096}{-4}=43524$

 

[kaliprasad]

my solution:

Spoiler

for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}=572724$

Spoiler

I am unable to derive /understand (4)

 

[Albert1]

Spoiler

I am unable to derive /understand (4)

Spoiler

$f(x)=x^3+20x-17$
$f(r)=r^3+20r-17=0---(1)$ ($r$ is a root of $f$)
$f(r+1)=(r+1)^3+20(r+1)-17=3r^2+3r+21$
from $(1):$ $r^3-1=(r-1)(r^2+r+1)=16-20r$
$r^2+r+1=\dfrac {16-20r}{r-1}$
$\therefore 3r^2+3r+21=\dfrac {30-42r}{r-1}$

 

[lfdahl]

my solution:

Spoiler

for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}=572724$

Hi, Albert!Thankyou very much for your interesting approach, but ...

Spoiler

- there must be a miscalculation somewhere. :( The correct end value is $43524$.

 

[Albert1]

Hi, Albert!Thankyou very much for your interesting approach, but ...

Spoiler

- there must be a miscalculation somewhere. :( The correct end value is $43524$.

Spoiler

yes a miscalculation found the answer is 43524
the solution has been edited
what a shame ! my poor calculation

 

[lfdahl]

Thankyou, Albert! - for your correct solution!
Please don´t bother because of the previous small miscalculation.
You are indeed a master in solving challenges and puzzles!

An alternative solution:

Spoiler

First we note, that $f(r_1+1) = 3r_1^2+3r_1+21$, since $r_1^3+20r_1-17 = 0$.

The quadratic polynomium has the roots:

\[3\left ( r_1^2+r_1+7 \right ) =3 \left ( \frac{-1+i3\sqrt{3}}{2}-r_1 \right )\cdot \left (\frac{-1-i3\sqrt{3}}{2}-r_1 \right ) = 3(\alpha -r_1)(\beta -r_1).\]

Now,
\[f(\alpha )=(\alpha-r_1)( \alpha-r_2)( \alpha-r_3 )\\\\
f(\beta )=(\beta-r_1)( \beta-r_2)( \beta-r_3 ) \\\\ \Rightarrow 3^3\cdot f(\alpha )\cdot f(\beta )=3(\alpha -r_1)(\beta -r_1) \cdot 3(\alpha -r_2)(\beta -r_2) \cdot 3(\alpha -r_3)(\beta -r_3)
\\\\=f(r_1+1)f(r_2+1)f(r_3+1)\]

\[f(\alpha )= -17+i 21\sqrt{3}, \: \: f(\beta )= -17-i 21\sqrt{3}\]

Thus, we get:

$ f(r_1+1)f(r_2+1)f(r_3+1)= 27f(\alpha ) f(\beta )= 27(17^2+3\cdot 21^2) = 43524$.

 

[Albert1]

Thankyou, Albert! - for your correct solution!
Please don´t bother because of the previous small miscalculation.
You are indeed a master in solving challenges and puzzles!

An alternative solution:

Spoiler

First we note, that $f(r_1+1) = 3r_1^2+3r_1+21$, since $r_1^3+20r_1-17 = 0$.

The quadratic polynomium has the roots:

\[3\left ( r_1^2+r_1+7 \right ) =3 \left ( \frac{-1+i3\sqrt{3}}{2}-r_1 \right )\cdot \left (\frac{-1-i3\sqrt{3}}{2}-r_1 \right ) = 3(\alpha -r_1)(\beta -r_1).\]

Now,
\[f(\alpha )=(\alpha-r_1)( \alpha-r_2)( \alpha-r_3 )\\\\
f(\beta )=(\beta-r_1)( \beta-r_2)( \beta-r_3 ) \\\\ \Rightarrow 3^3\cdot f(\alpha )\cdot f(\beta )=3(\alpha -r_1)(\beta -r_1) \cdot 3(\alpha -r_2)(\beta -r_2) \cdot 3(\alpha -r_3)(\beta -r_3)
\\\\=f(r_1+1)f(r_2+1)f(r_3+1)\]

\[f(\alpha )= -17+i 21\sqrt{3}, \: \: f(\beta )= -17-i 21\sqrt{3}\]

Thus, we get:

$ f(r_1+1)f(r_2+1)f(r_3+1)= 27f(\alpha ) f(\beta )= 27(17^2+3\cdot 21^2) = 43524$.

Spoiler

to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming
do you have a better way to get them ?

 

[lfdahl]

Spoiler

to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming
do you have a better way to get them ?

Spoiler

I´m afraid, no. Maybe, someone in the forum knows a less laborious way to follow?
The conversion to polar notation doesn´t make things better:

\[\frac{-1+i3\sqrt{3}}{2} \approx \sqrt{7}\cdot e^{i100.893^{\circ}}\]

 

[Greg]

Hi guys. Just a friendly reminder to use spoiler tags. Thanks. :D

 

[lfdahl]

Hi guys. Just a friendly reminder to use spoiler tags. Thanks. :D

Thankyou, greg1313, for the friendly reminder!
I will take precaution in future challenge & puzzle correspondence.