Question-211330

Question Number 211330 by RojaTaniya last updated on 06/Sep/24

Answered by mr W last updated on 06/Sep/24

s a y p = \frac{a}{b}, q = \frac{b}{c}, r = \frac{c}{a}

p q r = 1

p + q + r = 7

\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 11

\Rightarrow \frac{p q + q r + r p}{p q r} = 11 \Rightarrow p q + q r + r p = 11

{(p + q + r)}^{3} = p^{3} + q^{3} + r^{3} + 3 (p + q + r) (p q + q r + r p) - 3 p q r

7^{3} = p^{3} + q^{3} + r^{3} + 3 \times 7 \times 11 - 3 \times 1

\Rightarrow p^{3} + q^{3} + r^{3} = 7^{3} - 3 \times 7 \times 11 + 3 \times 1 = 115

\Rightarrow \frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} = 115

Commented by efronzo1 last updated on 06/Sep/24

how to get pqr = 1 sir

Commented by som(math1967) last updated on 06/Sep/24

p q r = \frac{a}{b} \times \frac{b}{c} \times \frac{c}{a} = 1

Commented by Tawa11 last updated on 06/Sep/24

Sir mrW,

Someone asked question 211340 and answered .

But I reposted because I want to see your approach sir .

I love to see your approach to question I like to learn .

Thanks sir .

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