# Find the value of the equation?

If a+b+c=1 and 1/a + 1/b + 1/c = 2 find the value of:

### 5 Answers

- JOHNLv 74 weeks agoFavorite Answer
EDIT: A - sign in line 8 should have been a + sign; now corrected.

Let 1/a , 1/b . 1/c be the roots of x³ + ux² + vx + w = 0

So –u = 1/a + 1/b + 1/c = 2 , u = -2

a, b, c are the roots of wy³ + vy² + uy + 1 = 0

a + b + c = 1 = -v/w, w = -v

So 1/a , 1/b . 1/c are the roots of

x³ - 2x² + vx - v = 0, abc = 1/v

(1/a – 1)(1/b – 1)(1/c – 1)

= 1/(abc) + (1/a + 1/b + 1/c) - (1/cb + 1/ ab + 1/ ca) – 1 = 1/(abc) + (1/a + 1/b + 1/c) – (1/(abc))(a + b + c) – 1 = v + 2 – v(1) – 1 = v + 2 – v - 1 = 1.

- MyRankLv 64 weeks ago
a+b+c = 1

1/a + 1/b + 1/c = 2(1/a - 1) (1/b - 1) (1/c - 1)1/a + 1/b + 1/c = 2bc + ac + ab / abc = 2bc + ac + ab = 2abc → bc + ac + ab / abc = 2.

Source(s): http://myrank.co.in/ - KrishnamurthyLv 74 weeks ago
If a + b + c = 1 and 1/a + 1/b + 1/c = 2

find the value of the equation (1/a - 1)(1/b - 1)(1/c - 1)

- Φ² = Φ+1Lv 74 weeks ago
Ground work: Chosing a value for a, find values for b+c and bc:

a = ⅔ → b + c = 1 - ⅔ = ⅓ → b, c = ⅙(1±√d) → bc = (1-d)/36.

1/b + 1/c = 2 - ³∕₂ = ½ → c + b = ½bc → ⅓ = ½(1-d)/36 → d = -23 → bc = ⅔

a = ⅔, b = ⅙(1+i√23), c = ⅙(1-i√23)

Checking:a + b + c = 1 ✓ and 1/a + 1/b + 1/c = 2 ✓

Manipulate and substitute for a, b+c and bc:

(1/a - 1)(1/b - 1)(1/c - 1)

= (1-a)(1-b)(1-c)/(abc)

= (1 - ⅔)(1 - (b + c) + bc)/(⅔bc)

= (⅓)(1 - ⅓ + ⅔)/(⅔⅔)

= (⁴∕₉)/(⁴∕₉)

= 1

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- 4 weeks ago
(1/a - 1) * (1/b - 1) * (1/c - 1) =>

((1 - a) / a) * ((1 - b) / b) * ((1 - c) / c) =>

(1 - a) * (1 - b) * (1 - c) / (a * b * c)

1/a + 1/b + 1/c = 2

(bc + ac + ab) / (a * b * c) = 2

(ab + ac + bc) / (a * b * c) = 2

(1 - a) * (1 - b) * (1 - c) / (a * b * c) =>

(a + b + c - a) * (a + b + c - b) * (a + b + c - c) / (abc) =>

(b + c) * (a + c) * (a + b) / (abc) =>

(a^2 + ab + ac + bc) * (b + c) / (abc) =>

(a^2 * (b + c)) / (abc) + (b + c) * (ab + ac + bc) / (abc) =>

(a^2 * (b + c)) / (abc) + (b + c) * 2 =>

(a^2 * b + a^2 * c + a * bc - a * bc) / (abc) + 2 * (b + c) =>

a * (ab + ac + bc) / (abc) - abc/(abc) + 2 * (b + c) =>

a * 2 - 1 + 2b + 2c =>

2a + 2b + 2c - 1 =>

2 * (a + b + c) - 1 =>

2 * 1 - 1 =>

2 - 1 =>

1

I followed this up to line 7 from the bottom. Then an abc seems to have disappeared. I make line 6 from the bottom 2a*/(abc) -1 + 2b + 2c. But I may be missing something.

But you have done something interesting and very useful to me, as it proved that my answer was wrong and led me to check my solution, where I discovered that I had a - sign where I should have has a + sign.