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Question Number 216312 by MATHEMATICSAM last updated on 03/Feb/25

If {a b}^{2} + {b c}^{2} + {c a}^{2} = 0 then find

(\frac{a}{b} + \frac{b}{c}) + (\frac{b}{c} + \frac{c}{a}) + (\frac{c}{a} + \frac{a}{b}) + 2 .

Answered by Rasheed.Sindhi last updated on 03/Feb/25

{a b}^{2} + {b c}^{2} + {c a}^{2} = 0

\frac{{a b}^{2}}{a b c} + \frac{{b c}^{2}}{a b c} + \frac{{c a}^{2}}{a b c} = 0

\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 0

2 (\frac{b}{c} + \frac{c}{a} + \frac{a}{b}) = 0

\frac{b}{c} + \frac{c}{a} + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 0

(\frac{a}{b} + \frac{b}{c}) + (\frac{b}{c} + \frac{c}{a}) + (\frac{c}{a} + \frac{a}{b}) + 2 = 2

Answered by som(math1967) last updated on 03/Feb/25

\frac{2 a}{b} + \frac{2 b}{c} + \frac{2 c}{a} + 2

= \frac{2 a^{2} c + 2 b^{2} a + 2 c^{2} b}{a b c} + 2

= \frac{2 ({a b}^{2} + {b c}^{2} + {c a}^{2})}{a b c} + 2

= \frac{2 \times 0}{a b c} + 2 = 2