Question 1046033
.
Please help me solving it 
If {{{a=(1/2) (5-21^(1/2))}}} then the value of {{{a^3+a^(-3)-5a^2-5a^(-2)+a+a^(-1)}}} is ?
~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
First, let us prove this statement: 


    if  a = {{{(1/2)*(5-sqrt(21))}}}  then  {{{a^(-1)}}} = {{{1/a}}} = {{{(1/2)*(5+sqrt(21))}}}.


Indeed,  {{{(1/2)*(5-sqrt(21))}}}*{{{(1/2)*(5+sqrt(21))}}} = {{{(1/4)*(5^2 - (sqrt(21)^2))}}} = {{{(1/4)*(25-21)}}} = {{{(1/4)*4}}} = 1.


OK. So, if  a = {{{(1/2)*(5-sqrt(21))}}},  then  {{{a + a^(-1)}}} = {{{(1/2)*(5-sqrt(21) + 5 + sqrt(21))}}} = {{{10/2}}} = 5.


Having this, we can make the next step.

{{{a^2}}} + {{{1/a^2}}} = {{{(a + 1/a)^2 - 2}}} = {{{5^2 - 2}}} = 23.   (I replaced here  {{{(a + 1/a)}}}  by 5, based on the proof above )


Furthermore, 

{{{a^3}}} + {{{1/a^3}}} = {{{(a+ 1/a)*(a^2 - 1 + 1/a^2)}}}          ( I applied the formula {{{x^3 + y^3}}} = {{{(x+y)*(x^2-xy+y^2)}}}

            = {{{5*(23-1)}}} = 110.             ( I replaced here {{{a + 1/a}}} by 5  and replaced  {{{a^2 - 1}}} + {{{1/a^2)}}}  by (23-1) = 22 based on results proved above).

Thus we have  {{{a + 1/a}}} = 5,  {{{a^2}}} + {{{1/a^2}}} = 23  and  {{{a^3}}} + {{{1/a^3}}} = 110.


Then the value under the question is  110 - 5*23 + 5 = 0.
</pre>

For similar problems, see 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, 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>.