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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} \frac{-b+ \sqrt{D} }{2a} .and. \frac{-b- \sqrt{D} }{2a}](https://tex.z-dn.net/?f=+%5Cfrac%7B-b%2B+%5Csqrt%7BD%7D+%7D%7B2a%7D+.and.+%5Cfrac%7B-b-+%5Csqrt%7BD%7D+%7D%7B2a%7D+)
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
______________________________________________________________
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
____________________________________________________________
Solution :
Here The quadratic equation have two roots
Where
a = c² - ab
b = -2(a² - bc)
c = b² - ac
D= b² - 4ac
But here D = b² - 4ac = 0
_________________________________________________________
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
___________________________________________________________
That is a = 0 OR a³ + b³ + c³ = 3abc
mysticd:
last line is wrong.plz edit
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