Question
If x = 8.15, y = 9.06 and z = –17.21, then the value
of x³ + y³ + z³ – 3xyz is:Solution
x³ + y³ + z³ – 3xyz = (x+ y + z)/2[(x-y) ² +(y-z) ² +(z-a) ²] = {8.15+9.06+(-17.21)}/2[(x-y) +(y-z) +(z-a)] = (17.21 -17.21)/2[(x-y) +(y-z) +(z-a)] =0[(x-y) +(y-z) +(z-a)] =0
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