Question

if roots of equation ( c2-ab)x2-2(a2-bc)x+b2-ac=0 are equal . showthat either a=0 or a3+b3+c3=3abc

Answer 1

$\bold{a=0 \text { or } a^{3}+b^{3}+c^{3}=3 a b c}$ is proven

Solution:

Let us consider a quadratic polynomial $p x^{2}+q x+r=0$  

We know that if discriminant $q^{2}-4 p r=0$, the roots of the equation are equal.

Here $q=-2\left(a^{2}-b c\right)$ ; $p=c^{2}-a b$ ; $r=b^{2}-a c$  

Therefore, discriminant=0

$\begin{array}{l}{q^{2}=4 p r} \\ \\{\left(-2\left(a^{2}-b c\right)\right)^{2}=4\left(c^{2}-a b\right)\left(b^{2}-a c\right)} \\ \\{4\left(a^{4}+b^{2} c^{2}-2 a^{2} b c\right)=4\left(b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}\right)} \\ \\{a^{4}+b^{2} c^{2}-2 a^{2} b c=b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}}\end{array}$

$\begin{array}{l}{a^{4}+b^{2} c^{2}-2 a^{2} b c-b^{2} c^{2}-a^{2} b c+a b^{3}+a c^{3}=0} \\ \\{a\left(a^{3}+b^{3}+c^{3}\right)=3 a^{2} b c} \\ \\{a\left(a^{3}+b^{3}+c^{3}-3 a b c\right)=0}\end{array}$

$\begin{array}{l}{a=0 \text { or } a^{3}+b^{3}+c^{3}-3 a b c=0} \\ \\{a=0 \text { or } a^{3}+b^{3}+c^{3}=3 a b c}\end{array}$

Hence, proved.

Answer 2

Answer(Explanation):

Compare given equation with ,

Ax² + Bx + C = 0 ,

A = c² - ab ;

B = - 2( a² - bc ) ;

C = b² - ac ;

It is given that , roots of the equation are equal ,

Therefore , Discriminant = 0

=>B² - 4AC = 0

=>[-2(a² - bc)]² - 4(c² - ab ) (b² -ac ) = 0

=>4(a⁴-2a²bc+b²c²)-4(b²c²-ac³-ab³+a²bc)=0

=>4[a⁴-2a² bc+b²c² - b² c² +ac³+ab³-a²bc ]=0

=>a⁴ - 3a²bc + ac³ + ab³ = 0

=>a( a³ + b³ + c³ - 3abc ) = 0

Therefore ,

a = 0 or a³ + b³ + c³ - 3abc = 0

a = 0 or a³ + b³ + c³ = 3abc

Hence proved.

Hope it helps!

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