Question

Find a all real numbers satisfying x^8+y^8=8xy-6?

Answer 1

Given :    x⁸  + y⁸   =  8xy  - 6  

To find :    all real numbers satisfying the given Equation

Solution:

x⁸  + y⁸   =  8xy  - 6

=> x⁸  + y⁸  + 6  =  8xy

=> x⁸  + y⁸  + 1 + 1 +1 + 1  + 1 + 1   =  8xy

=> (x⁸  + y⁸  + 1 + 1 +1 + 1  + 1 + 1 )/8  =  xy

AM of x⁸  + y⁸  + 1 + 1 +1 + 1  + 1 + 1  =   (x⁸  + y⁸  + 1 + 1 +1 + 1  + 1 + 1 )/8   = xy

GM of x⁸  + y⁸  + 1 + 1 +1 + 1  + 1 + 1   = (x⁸.y⁸.1.1.1.1.1.1)^(1/8) = xy

AM = GM  = xy  only possible

iff  all terms are Equal

=> x⁸ = y⁸ = 1

=> real value value of x & y = ± 1

Hence There are two solutions satisfying this

x = 1  , y = 1

LHS = 1⁸ + 1⁸  = 2

RHS = 8(1)(1) - 6  = 2

LHS = RHS

x = - 1 , y = -1

LHS = (-1)⁸ + (-1)⁸  = 2

RHS = 8(-1)(-1) - 6  = 2

LHS = RHS

x = 1  , y = 1    and x = - 1 , y = -1  

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Answer 2

Given: The equation x^8 + y^8 = 8xy - 6

To find: All real numbers satisfying the equation?

Solution:

• Now we have provided with the equation x^8 + y^8 = 8xy - 6.

            x and y must be of same sign else LHS > 0 and RHS < 0.

• We know that Arithmetic Mean is always greater that Geometric Mean. So:

            AM ≥ GM

            x^8 + y^8 + 6 = 8xy

            x^8 + y^8 + 1 + 1+ 1 + 1 + 1 + 1  ≥ 8 . (x^8 y^8)^1/8

            x^8 + y^8 + 1 + 1+ 1 + 1 + 1 + 1  ≥ 8 . |xy|

• So by hypothesis, this equality holds.

• All the eight terms are equal.

• So therefore x^8 = y^8 = 1

• Hence (x,y) = (1,1) or (-1,-1)

Answer:

         Hence (x,y) = (1,1) or (-1,-1) are the solution set for x^8 + y^8 = 8xy - 6