x+y+z=8 and xy+yz+zx=20 find x³+y³+z³-3xyz
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4
Answer: 32
Step-by-step explanation:
(x+y+z)³=x³+y³+z³+3{(x+y+z)(xy+yz+zx)-xyz}
= 8³ =x³+y³+z³+3(8*20-xyz)
= 512= x³+y³+z³+3(160-xyz)
= 512 =x³ +y³ +z³ +480-3xyz
= x³+y³+z³ -3xyz = 512 -480
Hence, x³+y³+z³-3xyz =32
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3
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