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Albert
Well-known member
- Jan 25, 2013
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$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$
$find:a^3+b^3+c^3=?$
$find:a^3+b^3+c^3=?$
$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$
$find:a^3+b^3+c^3=?$
good solutionLet 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