If-1-a-1-b-1-c-1-a-b-c-then-prove-that-1-a-3-1-b-3-1-c-3-1-a-b-c-3-

Question Number 173157 by AgniMath last updated on 07/Jul/22

If \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a + b + c} then prove that

\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}} = \frac{1}{{(a + b + c)}^{3}} .

Answered by Cesar1994 last updated on 07/Jul/22

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a + b + c}

{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})}^{3} = {(\frac{1}{a + b + c})}^{3}

\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}} + 3 (\frac{1}{a} + \frac{1}{b}) (\frac{1}{a} + \frac{1}{c}) (\frac{1}{b} + \frac{1}{c}) = \frac{1}{{(a + b + c)}^{3}} \cdot \cdot \cdot (1)

demostraremos que (\frac{1}{a} + \frac{1}{b}) (\frac{1}{a} + \frac{1}{c}) (\frac{1}{b} + \frac{1}{c}) = 0

Sea x = \frac{1}{a + b + c} \cdot \cdot \cdot (2)

\Rightarrow (\frac{1}{a} + \frac{1}{b}) (\frac{1}{a} + \frac{1}{c}) (\frac{1}{b} + \frac{1}{c}) = (x - \frac{1}{c}) (x - \frac{1}{b}) (x - \frac{1}{a})

= x^{3} - (\underset{x}{\underset{⏟}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}) x^{2} + (\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}) x - \frac{1}{a b c}

= x^{3} - x^{3} + (\frac{a + b + c}{a b c}) x - \frac{1}{a b c}, (a + b + c) x = 1 por (2)

= \frac{1}{a b c} - \frac{1}{a b c}

= 0

sustituyendo en (1) se concluye que

\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}} = \frac{1}{{(a + b + c)}^{3}}

Commented by Tawa11 last updated on 11/Jul/22

Great sirs

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