# Find the value of the equation?

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

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.

• 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.

• 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)

• 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

• JOHN
Lv 7
4 weeks agoReport

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.

• (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

• JOHN
Lv 7
4 weeks agoReport

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.