Question

If a + b + c = 18, a2 + b2 + c2 = 110 and a3 + b3 + c3 = 684, Find the value of (1/a + 1/b + 1/c)?
1. 13/17
2. 130/210
3. 107/210
4. 2/3

Answer 1

Correct Answer - Option 3 : 107/210

Given:

a + b + c = 18

a2 + b2 + c2 = 110

a3 + b3 + c3 = 684

Formula used:

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

a3 + b3 + c3 - 3abc = (a + b + c){a2 + b2 + c2 - (ab + bc + ca)}

Calculation:

First simplify what is asked

⇒ (1/a + 1/b + 1/c)

⇒ (ab + bc + ca)/abc      ----(1)

Now, find (ab + bc + ca)

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

⇒ (18)2 = 110 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 324 - 110

⇒ ab + bc + ca = 214/2 = 107

Put value in 2^nd formula

⇒ a3 + b3 + c3 - 3abc = (a + b + c){a2 + b2 + c2 - (ab + bc + ca)}

⇒ 684 - 3abc = (18)(110 - 107)

⇒ 3abc = 684 - 54

⇒ 3abc = 630

⇒ abc = 210

Put the values in the equation (1)

⇒ 107/210 

∴ The value of (1/a + 1/b + 1/c) is 107/210.