How can you solve for a^3 + b^3 + c^3?

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The discussion focuses on solving the equation $a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$ to find the value of $a^3+b^3+c^3$. Participants emphasize the importance of algebraic manipulation and substitution techniques to simplify the expression. The consensus is that understanding the properties of cubic equations and square roots is essential for deriving the solution effectively.

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$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$$find:a^3+b^3+c^3=?$
 
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Re: find :a^3+b^3+c^3

Albert said:
$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$$find:a^3+b^3+c^3=?$
Let a = $x^2$, b = $y^2-1$, c = $z^2+ 1$
We get $ x^2 + y^2 + z^2 + 3 = 2x + 2y + 2z$
Or$ (x^2 – 2x + 1) + (y^2 – 2y + 1) + (z^2-2z +1)=0$
Or $(x-1)^2 + (y-1)^2 + (z-1)^2 = 0$
x = y = z = 1 or a = 1, b = 0, c = 2 => $a^3 + b^3 + c^3$ = 9
 
Re: find :a^3+b^3+c^3

kaliprasad said:
Let a = $x^2$, b = $y^2-1$, c = $z^2+ 1$
We get $ x^2 + y^2 + z^2 + 3 = 2x + 2y + 2z$
Or$ (x^2 – 2x + 1) + (y^2 – 2y + 1) + (z^2-2z +1)=0$
Or $(x-1)^2 + (y-1)^2 + (z-1)^2 = 0$
x = y = z = 1 or a = 1, b = 0, c = 2 => $a^3 + b^3 + c^3$ = 9

good solution (Clapping)
 

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