Question

if roots to the equation

$( {c}^{2} - ab) {x}^{2} - 2( {a}^{2} - bc)x + {b}^{2} - ac = 0 \: \: ar e \: equal \\ \\ then \: prove \: that \: either \: a = 0 \: or \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

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BEST ANSWER WILL BE MARKED AS BRAINLIEST!

Answer 1

$<b><font color = "red" >Hello Friend$



⬇⬇Find your answer below⬇⬇



$\mathsf{\implies Given, Equation \: is:}$


$\mathsf{\implies (c^{2} - ab)x^{2} - 2(a^{2} - bc)x + (b^{2} - ac) = 0}$


$\mathsf{\implies To \: Prove : a = 0 \: or \: a^{ 3} + b^{3} + c^{3} = 3abc}$


$\mathsf {\implies Proof : From \: the \: given \: equation, we \: have}$


$\mathsf{\implies a = c^{2} - ab, \: b = - 2(a^{2} - bc), \: c = b ^{2} - ac}$


$\textsf {It is being given that equation has real and equal roots}$


$\mathsf {∴Delta = 0}$


$\mathsf{\implies b^{2} - 4ac = 0}$


$\textsf {On substituting respective values of a, b and c in above equation, we get}$


$\mathsf{\implies [- 2 (a^{2} - bc)]^{2} - 4(c^{2} - ab)(b^{2} - ac) = 0}$



$\mathsf {\implies 4(a^{2} - bc)^{2} - 4(c^{2}b^{2} - ac^{3} - ab^{3} + a^{2}bc) = 0}$



$\mathsf{\implies 4 (a^{4} + b^{2} c^{2} - 2a^{2} bc) - 4(c^{2}b ^{2} - ac^{3} - ab^{3} + a^{2}bc) = 0}$



$\mathsf{\implies 4a^{4} + 4b^{2}c^{2} - 8a^{2}bc - 4c^{2}b^{2} + 4ac^{3} + 4ab^{3} - 4a^{2}bc = 0}$



$\mathsf{\implies 4a^{4} - 12a^{2}bc + 4ac^{3} + 4ab^{3} = 0}$



$\mathsf{\implies 4a (a^{3} + b^{3} + c^{3} - 3abc) = 0}$



$\mathsf{\implies 4a = 0}$



$\mathsf{\implies a = 0}$



$\mathsf{\implies a^{3} + b^{3} + c^{3} - 3abc = 0}$



$<big><b>Hence Proved$



#BE BRAINLY



$<b><marquee>Thank you$

Answer 2

I think your question needs a correction.

To Prove : a = 0 Or a^3 + b^3 + c^3 = 3abc

Given equation : ( c^2 - ab )x^2 - 2( a^2 - bc )x + ( b^2 - ac ) = 0

On comparing the given equation with ax^2 + bx + c = 0, we get the following information :

a = ( c^2 - ab ) , b = - 2( a^2 - bc ) , c = ( b^2 - ac )

⇒ Discriminant = 0

⇒ b^2 - 4ac = 0

⇒ [ -2( a^2 - bc ) ]^2 - 4( c^2 - ab )( b^2 - ac ) = 0

⇒ 4[ a^4 + b^2 c^2 - 2a^2 bc ] - 4[ c^2 b^2 - ac^3 - ab^3 + bca^2 ] = 0

⇒ [ a^4 + b^2 c^2 - 2a^2 bc ] - [ c^2 b^2 - ac^3 - ab^3 + bca^2 ] = 0

⇒ a^4 + b^2 c^2 - 2a^2 bc - c^2 b^2 + ac^3 + ab^3 - bca^2 =0  

⇒ a^4 - 2a^2 bc + ac^3 + ab^3 - bca^2 = 0

⇒ ( a )(  a^3 - 2abc + c^3 + b^3 - abc ) = 0

⇒ ( a )( a^3 - 2abc + c^3 + b^3 - abc ) = 0

⇒ ( a ) ( a^3 + b^3 + c^3 - 3abc ) = 0

             By Zero Product Rule

⇒ a^3 + b^3 + c^3 - 3abc = 0

Or a = 0

⇒ a^3 + b^3 + c^3 = 3abc

Or , a = 0

Hence, proved