if a+b+c=(1/(a+1))+(1/(b+2))+(1/(c+3))=0, find (a+1)^2 +(b+2)^2 +(c+3)^2 =?


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Question Number 205626 by mr W last updated on 25/Mar/24

$i f a + b + c = \frac{1}{a + 1} + \frac{1}{b + 2} + \frac{1}{c + 3} = 0,$ $f i n d {(a + 1)}^{2} + {(b + 2)}^{2} + {(c + 3)}^{2} = ?$
	Answered by A5T last updated on 25/Mar/24

	$a + b + c = 0 \land (b + 2) (c + 3) + (a + 1) (b + 2) + (a + 1) (c + 3) = 0$ ${(a + 1)}^{2} + {(b + 2)}^{2} + {(c + 3)}^{2}$ $= {(a + 1 + b + 2 + c + 3)}^{2} - 2 (a + 1) (b + 2) - 2 (a + 1) (c + 3)$ $- 2 (c + 3) (b + 2) = 6^{2} = 36$
	Commented by mr W last updated on 26/Mar/24
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	Answered by mr W last updated on 26/Mar/24

	$s a y x = a + 1, y = b + 2, z = c + 3$ $\Rightarrow x + y + z = 1 + 2 + 3 = 6$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \Rightarrow x y + y z + z x = 0$ ${(a + 1)}^{2} + {(b + 2)}^{2} + {(c + 3)}^{2}$ $= x^{2} + y^{2} + z^{2}$ $= {(x + y + z)}^{2} - 2 (x y + y z + z x)$ $= 6^{2} - 2 \times 0 = 36 ✓$
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