Question 1122043
.
<pre>
You are given


{{{a}}} - {{{1/a}}} = 3.


Square both sides.  You will get


9 = {{{(a - 1/a)^2}}} = {{{a^2}}} - {{{2a*(1/a)}}} + {{{1/a^2}}} = {{{a^2}}} - 2 + {{{1/a^2}}}.


Hence,  {{{a^2}}} + {{{1/a^2}}} = 9 + 2 = 11.


Then    {{{a^2}}} + {{{3a}}} + {{{1/a^2}}} - {{{3/a}}}  (which is under the question) is equal to


    {{{(a^2 + 1/a^2)}}} + {{{3*(a-1/a)}}} = 11 +3*3 = 20.


<U>Answer</U>.  Then  {{{(a^2 + 1/a^2)}}} + {{{3*(a-1/a)}}} = 20.
</pre>

Solved.


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


To see many other similar solved problems, look into the lesson

&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>

in this site.


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 lesson is 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.