If a,b and c are the roots of x^3 −21x−35 = 0 what is the value of (a−b)(c−a)(b−c) ?


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Question Number 129336 by bramlexs22 last updated on 15/Jan/21

$If a, b and c are the roots of$ $x^{3} - 21 x - 35 = 0 what is the value of$ $(a - b) (c - a) (b - c) ?$
Commented by MJS_new last updated on 22/Jan/21

$x^{3} + p x + q = 0$ $Cardano gives$ $u = - \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}} \land v = - \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}$ $[and for this purpose they are allowed to be$ $complex]$ $the roots then are [ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i]$ $a = u^{1 / 3} + v^{1 / 3}$ $b = ω u^{1 / 3} + w^{2} v^{1 / 3}$ $c = ω^{2} u^{1 / 3} + ω v^{1 / 3}$ $(a - b) (b - c) (c - a) =$ $[knowing that ω^{3} = 1, ω^{3 n + k} = ω^{k}]$ $. . .$ $= 3 (ω^{2} - ω) (u - v) =$ $= - 3 \sqrt{3} (u - v) i =$ $= - i \sqrt{4 p^{3} + 27 q^{2}}$ $with p = - 21 \land q = - 35 we get 63$
Commented by bramlexs22 last updated on 15/Jan/21

$correct sir . but how step to got it ?$
Commented by liberty last updated on 15/Jan/21
$my way. let p(x) = (x−a)(x−b)(x−c) be polynomial has the roots are a,b and c so we have (x−a)(x−b)(x−c)=x^3 −21x−35 differentiating both sides w.r.t x gives (x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=3x^2 −21 putting x=a⇒3a^2 −21=(a−b)(a−c) putting x=b⇒3b^2 −21=(b−a)(b−c) putting x=c⇒3c^2 −21=(c−a)(c−b) multiply three equation we get (a−b)(a−c)(b−a)(b−c)(c−a)(c−b)=27(a^2 −7)(b^2 −7)(c^2 −7) [ (a−b)(b−c)(c−a) ]^2 = 27(a^2 −7)(b^2 −7)(c^2 −7) now consider the original equation x^3 −21x = 35 ; x(x^2 −21)=35 x^2 (x^2 −21)^2 =35^2 ; let y=x^2 −7 (y+7)(y+7−21)^2 =35^2 ⇒y^3 −21y^2 −147=0 whose roots are { ((a^2 −7 )),((b^2 −7 )),((c^2 −7)) :} then by Vieta′s rule we get (a^2 −7)(b^2 −7)(c^2 −7)=147 finally we find [ (a−b)(b−c)(c−a) ]^2 =27×147 ⇔ (a−b)(b−c)(c−a) = ± (√(3969)) = ± 63$
$my way .$ $let p (x) = (x - a) (x - b) (x - c) be polynomial$ $has the roots are a, b and c so we have$ $(x - a) (x - b) (x - c) = x^{3} - 21 x - 35$ $differentiating both sides w . r . t x gives$ $(x - b) (x - c) + (x - a) (x - c) + (x - a) (x - b) = {3 x}^{2} - 21$ $putting x = a \Rightarrow {3 a}^{2} - 21 = (a - b) (a - c)$ $putting x = b \Rightarrow {3 b}^{2} - 21 = (b - a) (b - c)$ $putting x = c \Rightarrow {3 c}^{2} - 21 = (c - a) (c - b)$ $multiply three equation we get$ $(a - b) (a - c) (b - a) (b - c) (c - a) (c - b) = 27 (a^{2} - 7) (b^{2} - 7) (c^{2} - 7)$ ${[(a - b) (b - c) (c - a)]}^{2} = 27 (a^{2} - 7) (b^{2} - 7) (c^{2} - 7)$ $now consider the original equation$ $x^{3} - 21 x = 35; x (x^{2} - 21) = 35$ $x^{2} {(x^{2} - 21)}^{2} = 35^{2}; let y = x^{2} - 7$ $(y + 7) {(y + 7 - 21)}^{2} = 35^{2} \Rightarrow y^{3} - {21 y}^{2} - 147 = 0$ $whose roots are {\begin{cases} a^{2} - 7 \\ b^{2} - 7 \\ c^{2} - 7 \end{cases}$ $then by {Vieta}^{'} s rule we get (a^{2} - 7) (b^{2} - 7) (c^{2} - 7) = 147$ $finally we find {[(a - b) (b - c) (c - a)]}^{2} = 27 \times 147$ $\underset{―}{\Leftrightarrow (a - b) (b - c) (c - a) = \pm \sqrt{3969} = \pm 63}$
Commented by MJS_new last updated on 15/Jan/21

$I corrected and completed$
Commented by bramlexs22 last updated on 15/Jan/21

$waw . . . thank you$
Commented by bemath last updated on 15/Jan/21
$my method ⇔ x^3 −21x−35 = 0 a^3 −21a−35=0...(1) b^3 −21b−35=0...(2) (1)−(2)⇒a^3 −b^3 −21(a−b)=0 (a−b)(a^2 +ab+b^2 )−21(a−b)=0 (a−b)(a^2 +b^2 +ab−21)=0 ⇒(a−b)^2 +3ab−21=0 (a−b)^2 = 21−3ab similar to { (((c−a)^2 = 21−3ac)),(((b−c)^2 = 21−3bc)) :} so we find [(a−b)(c−a)(b−c)]^2 = (21−3ab)(21−3ac)(21−3bc) RHS ≡ −27(abc)^2 +189abc(a+b+c)− 1323(ab+ac+bc)+21^3 RHS≡ −27(35)^2 +189(0)−1323(−21)+21^3 RHS≡ 3969 finally we find (a−b)(c−a)(b−c) = ± (√(3969)) = ± 63$
$my method$ $\Leftrightarrow x^{3} - 21 x - 35 = 0$ $a^{3} - 21 a - 35 = 0 . . . (1)$ $b^{3} - 21 b - 35 = 0 . . . (2)$ $(1) - (2) \Rightarrow a^{3} - b^{3} - 21 (a - b) = 0$ $(a - b) (a^{2} + ab + b^{2}) - 21 (a - b) = 0$ $(a - b) (a^{2} + b^{2} + ab - 21) = 0$ $\Rightarrow {(a - b)}^{2} + 3 ab - 21 = 0$ ${(a - b)}^{2} = 21 - 3 ab$ $similar to {\begin{cases} {(c - a)}^{2} = 21 - 3 ac \\ {(b - c)}^{2} = 21 - 3 bc \end{cases}$ $so we find$ ${[(a - b) (c - a) (b - c)]}^{2} = (21 - 3 ab) (21 - 3 ac) (21 - 3 bc)$ $RHS \equiv - 27 {(abc)}^{2} + 189 abc (a + b + c) -$ $1323 (ab + ac + bc) + 21^{3}$ $RHS \equiv - 27 {(35)}^{2} + 189 (0) - 1323 (- 21) + 21^{3}$ $RHS \equiv 3969$ $finally we find (a - b) (c - a) (b - c) = \pm \sqrt{3969} = \pm 63$
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