Question

If the equations ax² + bx + c = 0 and cx² + bx + a = 0 has one root in common, then the value of a³ + b³ + c³ is -

(a) 3a(a + b + c)
(b) a² + b² + c² - ab - bc - ac
(c) 3(ab + bc + ac)
(d) 3abc​

Answer 1

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$\large \pmb{\sf{Question :-}}$

If the equations ax² + bx + c = 0 and cx² + bx + a = 0 has one root in common, then the value of a³ + b³ + c³ is -

(a) 3a(a + b + c)

(b) a² + b² + c² - ab - bc - ac

(c) 3(ab + bc + ac)

(d) 3abc

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$\large \pmb{\sf{Final \: Solution :-}}$

$\pink \bigstar \: \color{green} \boxed{ \sf\color{blue}{ : \leadsto {a}^{3} + {b}^{3} + {c}^{3} = 3abc }}$

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$\large \pmb{\sf{Explanation :-}}$

Let k be the common root of the equations ax² + bx + c = 0 and bx² + cx + a = 0

∴ak² + bk + c = 0

bk² + ck + a = 0

$\sf : \leadsto \frac{ {a}^{2} }{ab - {c}^{2} } = \frac{k}{bc - {a}^{2} } = \frac{1}{ac - {b}^{2} }$

$\sf: \leadsto k = \frac{bc - {a}^{2} }{ac - {b}^{2} }$

$\sf: \leadsto( cb - {a}^{2} )^{2} = (ab - {c}^{2} )(ac - {b}^{2} )$

$\sf: \leadsto {a}^{3} + {b}^{3} + { c}^{3} - 3abc = 0$

$\blue \bigstar \: \color{green} \boxed{\sf \color{maroon}: \leadsto {a}^{3} + {b}^{3} + { c}^{3} = 3abc }$

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$\sf \color{azure}\fcolorbox{pink}{purple}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \#Bebrainly \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\:\: \: \: \: \:}$

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$\sf \color{azure}\fcolorbox{pink}{black}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: @Saanvigrover2007 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

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Answer 2

this photo is your answer