SOLUTION: If x^4+y^4=2 and x^8+y^8=3, what is the value of x^(16)+y^(16)

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Question 1086719: If x^4+y^4=2 and x^8+y^8=3, what is the value of x^(16)+y^(16)
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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1.  If x%5E4+%2B+y%5E4 = 2 then (squaring both sides)

       x%5E8 + 2x%5E4%2Ay%5E4 + y%5E8 = 4.    (1)


    Using x%5E8%2By%5E8 = 3 (which is given) from (1) you get 2x%5E4%2Ay%5E4 = 1.    (2)


    Hence,  x%5E4%2Ay%5E4 = 1%2F2.


2.  Now, x%5E16+%2B+y%5E16 = %28x%5E8%2By%5E8%29%5E2+-+2x%5E8%2Ay%5E8 = 3%5E2+-+2%2A%28x%5E8%2Ay%5E8%29 = 9+-+2%2A%281%2F2%29%5E2 = 81%2F2.

Answer. x%5E16+%2B+y%5E16 = 81%2F2.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
If x^4+y^4=2 and x^8+y^8=3, what is the value of x^(16)+y^(16)
x%5E4+%2B+y%5E4+=+2 				  x%5E8+%2B+y%5E8+=+3
%28x%5E4+%2B+y%5E4%29%5E2+=+2%5E2 ------ Squaring both sides
x%5E8+%2B+2x%5E4y%5E4+%2B+y%5E8+=+4 
x%5E8+%2B+y%5E8+%2B+2x%5E4y%5E4+=+4 ------ Rearranging terms
3+%2B+2x%5E4y%5E4+=+4 ------ Substituting 3 for x%5E8+%2B+y%5E8 
2x%5E4y%5E4+=+1 
x%5E4y%5E4+=+1%2F2 ------ eq (i)


x%5E8+%2B+y%5E8+=+3+
%28x%5E8+%2B+y%5E8%29%5E2+=+3%5E2 ------ Squaring both sides
x%5E16+%2B+2x%5E8y%5E8+%2B+y%5E16+=+9 
x%5E16+%2B+y%5E16+=+9+-+2x%5E8y%5E8 
x%5E16+%2B+y%5E16+=+9+-+2%28x%5E4y%5E4%29%5E2
x%5E16+%2B+y%5E16+=+9+-+2%281%2F2%29%5E2 ------- Substituting ½ for x%5E4y%5E4 
x%5E16+%2B+y%5E16+=+9+-+2%281%2F4%29