If a + b + c = 1, a2+b2+c2=9, a3+b3+c3=1 find 1/a + 1/b + 1/c
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Answer:
1/a + 1/b + 1/c = 1
Step-by-step explanation:
If a + b + c = 1, a2+b2+c2=9, a3+b3+c3=1 find 1/a + 1/b + 1/c
a + b + c = 1
Squaring both sides
a² + b² + c² + 2(ab + bc + ca) = 1
=> 9 + 2(ab + bc + ca) = 1
=> 2(ab + bc + ca) = -8
=> ab + bc + ca = -4
a³+b³+c³ -3abc = (a² + b² + c² - (ab + bc + ca))(a + b + c)
=> 1 -3abc = (9 -(-4))(1)
=> 1 - 13 = 3abc
=> abc = -4
1/a + 1/b + 1/c
= (bc + ac + ab)/abc
= -4/(-4)
= 1
1/a + 1/b + 1/c = 1
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