document.write( "Question 882087: We are to find the value of X^4+Y^4+Z^4 when X, Y, Z are real numbers which satisfy the following three equalities:\r
\n" ); document.write( "\n" ); document.write( "X+Y+Z=3
\n" ); document.write( "X^2+Y^2+Z^2=9
\n" ); document.write( "XYZ=-2\r
\n" ); document.write( "\n" ); document.write( "Firstly it follows from the first 2 equalities that:
\n" ); document.write( "XY+YZ+ZX= A\r
\n" ); document.write( "\n" ); document.write( "Next using:
\n" ); document.write( "(X^2+Y^2+Z^2)^2=X^4+Y^4+Z^4+B ( (xy)^2+(yz)^2+(zx)^2 )\r
\n" ); document.write( "\n" ); document.write( "we have:
\n" ); document.write( " X^4+Y^4+Z^4= C
\n" ); document.write( "Find A, B, C
\n" ); document.write( "I tried to solve it by substituting by X in the first 3 equations but I can't seem to get a value for X or Y to solvr it please help \n" ); document.write( "

Algebra.Com's Answer #532697 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
X^4+Y^4+Z^4 = 57,
\n" ); document.write( "X = 1+sqrt(3),
\n" ); document.write( "Y = 1,
\n" ); document.write( "Z = 1-sqrt(3)\r
\n" ); document.write( "\n" ); document.write( "It doesn't matter which is which
\n" ); document.write( "x can be any of the three values
\n" ); document.write( "y can be any of the three
\n" ); document.write( "z can be any of the three
\n" ); document.write( "Those are the three values
\n" ); document.write( " \n" ); document.write( "

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