abc=8 a+b+c=7 a^3 +b^3 +c^3 =73 then faind the vole of (1/a)+(1/b)+(1/c)=?


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Question Number 161947 by mathlove last updated on 24/Dec/21

$a b c = 8$ $a + b + c = 7$ $a^{3} + b^{3} + c^{3} = 73$ $t h e n f a i n d t h e v o l e o f$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = ?$
	Answered by Rasheed.Sindhi last updated on 24/Dec/21

	$a b c = 8 \land a + b + c = 7 \land a^{3} + b^{3} + c^{3} = 73$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = ?$ $∙ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$ $a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 7^{2} = 49 . . . (i)$ $∙ a^{3} + b^{3} + c^{3} - 3 a b c = 73 - 3 (8) = 49$ $(a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a)) = 49$ $(7) (a^{2} + b^{2} + c^{2} - (a b + b c + c a)) = 49$ $a^{2} + b^{2} + c^{2} - (a b + b c + c a) = 7 . . . . . . . . (i i)$ $(i) - (i i) :$ $3 (a b + b c + c a) = 42$ $a b + b c + c a = 14$ $\frac{a b + b c + c a}{a b c} = \frac{14}{8} = \frac{7}{4}$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{7}{4}$
	Answered by mr W last updated on 24/Dec/21

	$a n o t h e r w a y :$ $p_{2} = e_{1} p_{1} - 2 e_{2} = 7 \times 7 - 2 e_{2} = 49 - 2 e_{2}$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$ $73 = 7 (49 - 2 e_{2}) - 7 e_{2} + 3 \times 8$ $\Rightarrow e_{2} = 14$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{a b + b c + c a}{a b c} = \frac{e_{2}}{e_{3}} = \frac{14}{8} = \frac{7}{4}$
	Commented by Tawa11 last updated on 24/Dec/21

	$Great sir$
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