if-a-b-c-1-a-1-1-b-2-1-c-3-0-find-a-1-2-b-2-2-c-3-2-

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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