Question

If the roots of equation (c2 - ab) x2 - 2(a2 -bc) x + b2 -ac = 0 are equal, prove that either a=o, or a3+b3+c3 = 3abc.

Answer 1

Answer:

Step-by-step explanation:

(c2 - ab) x2 - 2(a2 -bc) x + b2 -ac = 0

Here A=c2 - ab,B=- 2(a2 -bc),C=-ac

If the roots are equal

B^2=4A*C

[- 2(a2 -bc)]²=4*(c2-ab)(b2-ac)

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

a4+b2c2-2a2bc=-ac3+a2bc+b2c2-ab3

a4+ac3+ab3-2a2bc-a2bc=0

a(a3+b3+c3-3abc)=0

So a= 0 or

a3+b3+c3-3abc=0

a3+b3+c3=3abc