Question

IF X+Y+Z=0 , THEN FIND X^4+Y^4+Z^4=4XYZ

Answer 1

Answer:

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Step-by-step explanation:

The left side is equal to (x2−y2)2+2x2y2+z4−4xyz(x2−y2)2+2x2y2+z4−4xyz. The presence of 2x2y22x2y2 and −4xyz−4xyz suggests the possibility of adding a 2z22z2 and then completing one more square. So the equation can be rewritten as (x2−y2)2+(z2−1)2+2(xy−z)2=0(x2−y2)2+(z2−1)2+2(xy−z)2=0. This equality can hold only if all three squares are equal to zero. From z2−1=0z2−1=0 we have z=±1z=±1, and after a quick analysis we conclude that the solutions are 

Answer 2

Step-by-step explanation:

you can use identity: if a+b+c = 0, then a^3+ b^3 + c^3 = 3a*b*c.