Question

is a + b + c = 1, a² + b² + c² = 9, a³+ b³ + c³ = 1, then find 1/a + 1/b + 1/c​

Answer 1

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

Step-by-step explanation:

Answer 2

Step-by-step explanation:

this is your answer ......