SOLUTION: If x+y=4,xy=2 THEN X^6+Y^6=?

Question 1095680: If x+y=4,xy=2 THEN X^6+Y^6=?

Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Given:
x+y = 4
xy = 2

We'll call the equations shown above to be equation (1) and equation (2) (in that order)

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Multiply both sides of equation (2) by 2

xy = 2

2*xy = 2*2

2xy = 4

We'll refer to this as equation (3).

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

Square both sides of equation (1)

x+y = 4

(x+y)^2 = 4^2

x^2+2xy+y^2 = 16

x^2+4+y^2 = 16 ....note how the "2xy" term has been replaced with 4. This is valid due to equation (3)

x^2+4+y^2-4 = 12-4 ... subtract 4 from both sides

x^2+y^2 = 12

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We will refer to x^2+y^2 = 12 as equation (4).

Cube both sides of equation (4) to get the following:

x^2+y^2 = 12

(x^2+y^2)^3 = 12^3

(x^2)^3+3(x^2)^2(y^2)+3(x^2)(y^2)^2+(y^2)^3 = 1728

x^6+3x^4y^2+3x^2y^4+y^6 = 1728

(x^6+y^6)+3x^4y^2+3x^2y^4 = 1728

(x^6+y^6)+3x^2y^2(x^2+y^2) = 1728

(x^6+y^6)+3(xy)^2(x^2+y^2) = 1728

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Now make substitutions. We'll use equation (2) and equation (4)

(x^6+y^6)+3(xy)^2(x^2+y^2) = 1728

(x^6+y^6)+3(2)^2(x^2+y^2) = 1728 ... Use equation (2) to make the first substitution. Replace xy with 2.

(x^6+y^6)+3(2)^2(12) = 1728 ... Use equation (4) to make the next substitution. Replace x^2+y^2 with 12

(x^6+y^6)+144 = 1728

(x^6+y^6)+144-144 = 1728-144 ... Subtract 144 from both sides

x^6+y^6 = 1584

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Final Answer: 1584

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
This problem is solved in two steps:

Step 1. If x+y = 4 and xy = 2 then = = = 16 - 4 = 12. Step 2. Thus we proved that = 12. It implies further = = = . In the last expression, replace x^2*y^2 by = 4 and replace by 12. Then you can continue this chain of equalities in this way = = = = = . Thus = = 1584.

Answer.   . . . then = 1584.

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For similar solved problems, see the lessons
    - HOW TO evaluate expressions involving , and
    - Advanced lesson on evaluating expressions
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

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

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.

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