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Question Number 86855 by jagoll last updated on 01/Apr/20

If a, b, c are the roots of the equation

x^{3} + {6 x}^{2} - 4 x + 3 = 0 . find the

equation with roots a + b, b + c, a + c ?

Answered by john santu last updated on 01/Apr/20

by vieta

(i) a + b + c = - 6, ab + bc + ac = - 4

abc = - 3

we seek p, q, r where x^{3} + {px}^{2} + qx + r = 0

here we start

p = - (2 a + 2 b + 2 c) = 12

q = (a + b) (b + c) + (a + b) (a + c) + (b + c) (a + c)

q = (- 6 - c) (- 6 - a) + (- 6 - c) (- 6 - b) + (- 6 - a) (- 6 - b)

q = 36 \times 3 - 6 (2 a + 2 b + 2 c) + ab + ac + bc

q = 108 - 72 - 4 = 32

r = (a + b) (b + c) (a + c)

r = (- 6 - c) (- 6 - a) (- 6 - b)

r = - (216 + 36 (a + b + c) + 6 (ab + bc + ac) + abc)

r = - (216 + 36 (- 6) + 6 (- 4) - 3)

r = - 27

so the required equation is

x^{3} + {12 x}^{2} + 32 x - 27 = 0

Commented by MJS last updated on 01/Apr/20

it must be

x^{3} + 12 x^{2} + 32 x - 27 = 0

Commented by john santu last updated on 01/Apr/20

yes sir . i mistake calculate

Commented by john santu last updated on 01/Apr/20

why x = t - \frac{a}{3} ?

Commented by jagoll last updated on 01/Apr/20

it is cardano method sir

Commented by MJS last updated on 01/Apr/20

to eliminate the square factor

generally

x^{n} + c_{n - 1} x^{n - 1} + c_{n - 2} x^{n - 2} + \dots + c_{0} = 0

let x = t - \frac{c_{n - 1}}{n}

\Rightarrow

t^{n} + 0 t^{n - 1} + γ_{n - 2} t^{n - 2} + \dots + γ_{0} = 0

i . e .

x^{2} + c_{1} x + c_{0} = 0

x = t - \frac{c_{1}}{2}

t^{2} - \frac{c_{1}^{2}}{4} + c_{0} = 0

\Rightarrow t = \pm \sqrt{\frac{c_{1}^{2}}{4} - c_{0}}

leads to the well - known formula

x = - \frac{c_{1}}{2} \pm \sqrt{\frac{c_{1}^{2}}{4} - c_{0}}

Commented by MJS last updated on 01/Apr/20

another thing

x^{3} + {a x}^{2} + b x + c = 0

with solutions x = α, β, γ

let x = t - \frac{a}{3}

t^{3} - \frac{(a^{2} - 3 b)}{3} t + \frac{2 a^{3} - 9 a b + 27 c}{27} = 0

now change to

t^{3} - \frac{(a^{2} - 3 b)}{3} t - \frac{2 a^{3} - 9 a b + 27 c}{27} = 0

and let t = x + \frac{2 a}{3}

x^{3} + 2 {a x}^{2} + (a^{2} + b) x + a b - c = 0

will have the solutions α + β, α + γ, β + γ

can you prove it ?

Commented by john santu last updated on 01/Apr/20

waw \dots . super sir