Question 1095680
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This problem is solved in two steps:


<pre>
<U>Step 1</U>.

    If x+y = 4 and xy = 2 then  {{{x^2 + y^2}}} = {{{(x+y)^2 - 2xy}}} = {{{4^2 - 2*2}}} = 16 - 4 = 12.



<U>Step 2</U>. 

    Thus we proved that  {{{x^2 + y^2}}} = 12.  It implies further


    {{{(x^2 + y^2)^3}}} = {{{12^3}}} = {{{x^6 + 3x^4*y^2 + 3x^2*y^4 + y^6}}} = {{{x^6 + 3x^2*y^2*(x^2 + y^2) + y^6}}}.


    In the last expression, replace x^2*y^2 by {{{2^2}}} = 4  and  replace {{{x^2+y^2}}} by 12. 


    Then you can continue this chain of equalities in this way


    {{{(x^2 + y^2)^3}}} = {{{12^3}}} = {{{x^6 + 3x^4*y^2 + 3x^2*y^4 + y^6}}} = {{{x^6 + 3x^2*y^2*(x^2 + y^2) + y^6 }}} = {{{x^6 + 3*4*12 + y^6}}} = {{{x^6 + 144 + y^6}}}.


    Thus  {{{x^6+y^6}}} = {{{12^3 - 144}}} = 1584.
</pre>

<U>Answer</U>.   &nbsp;&nbsp;. . . then  &nbsp;{{{x^6+y^6}}} = 1584.



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For similar solved problems, 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>

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 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.