Question

Plz solve this question.

Answer 1

Hello Dearie @_@
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The original statement of this question is :

If the roots of equation (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 are equal . 

Prove that a³ + b³ + c³ = 3abc  Or  a = 0 

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Solution :

Here The quadratic equation have two roots 

$\frac{-b+ \sqrt{D} }{2a} .and. \frac{-b- \sqrt{D} }{2a}$

Where 
a = c² - ab
b = -2(a² - bc)
c = b² - ac
D= b² - 4ac
But here D = b² - 4ac = 0 
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i.e.[ -2(a² - bc)]² - 4 (c² - ab)(b² - ac)                       = 0 
       4(a² - bc)²  - 4 (c²b² - ac³ - ab³ + a²bc)             = 0
       4 (a⁴ + b²c² - 2a²bc) -4(c²b² - ac³ - ab³ + a²bc  = 0
       a⁴ + b²c² - 2a²bc - c²b² + ac³ + ab³ - a²bc         = 0
       a⁴ + ac³ + ab³ - 3a²bc                                        = 0
       a( a³ + b³ + c³ -3abc)                                         = 0
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That is a = 0 OR a³ + b³ + c³ = 3abc