Question 1172151
.
<pre>

According to Vieta's theorem, we have

   r + s = -3,  rs = -5.    (1)


Keeping it in mind, we can write


    {{{(-3)^3}}} = {{{(r+s)^3}}} = {{{r^3 + 3r^2s + 3rs^2 + s^3}}} = {{{r^3 + 3rs*(r+s) + s^3}}}.


Replace here  rs  by  -5  and replace  r+s  by  -3, based on (1).  You will get 

   -27 = {{{r^3 + s^3}}} + 3*(-5)*(-3) = {{{r^3 + s^3}}} + 45,


and then

    {{{r^3 + s^3}}} = -27 - 45 = -72.


<U>ANSWER</U>.   {{{r^3 + s^3}}} = -72.
</pre>

Solved.


--------------


See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-expressions-involving-x%2Binv%28x%29-x2%2Binv%28x2%29-and-x%5E3%2Binv%28x%5E3%29.lesson>HOW TO evaluate expressions involving &nbsp;{{{(x + 1/x)}}}, &nbsp;{{{(x^2+1/x^2)}}} &nbsp;and &nbsp;{{{(x^3+1/x^3)}}}</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-lesson-on-evaluating-expressions.lesson>Advanced lesson on evaluating expressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-square-equation.lesson>HOW TO evaluate functions of roots of a square equation</A>

in this site.


You will find there many similar problems solved.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Evaluation, substitution</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.