Question

If the roots of the quadratic equation `(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0` are real and equal, show that either  a=0  or  `(a^3+b^3+c^3=3abc)` 

 

Answer 1

Given: 

`(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0` 

Here, 

`a=(c^2-ab), b==-2(a^2-bc),c=(b^2-ac)=0` 

`D=0` 

⇒ `(b^2-4ac)=0` 

⇒` {-2(a^2-bc)}^2-4xx(c^2-ab)xx(b^2-ac)=0` 

⇒`4(a^4-2a^2bc+b^2c^2) -4(b^2c^2-ac^3-ab^3+a^2bc)=0` 

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

⇒`a^4-3a^2bc+ac^3+ab^3=0` 

⇒`a(a^3-3abc+c^3+b^3)=0`  

Now, 

`a= a^3-3abc+c^3+b^3=0`  

`a=0 or  a^3+b^3+c^3abc`