If ((u^5 +v^5 )/((u+v)^5 )) = −(1/5) , find ((u^3 +v^3 )/((u+v)^3 )) = ?


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Question Number 54607 by ajfour last updated on 07/Feb/19

$I f \frac{u^{5} + v^{5}}{{(u + v)}^{5}} = - \frac{1}{5},$ $f i n d \frac{u^{3} + v^{3}}{{(u + v)}^{3}} = ?$
Commented by MJS last updated on 08/Feb/19

$I think it can be - \frac{4}{5} or - \frac{1}{5}$
Commented by ajfour last updated on 08/Feb/19

$y e s s i r, i g e t - \frac{1}{5} .$
	Answered by mr W last updated on 08/Feb/19

	$l e t \frac{v}{u} = t$ $\frac{u^{5} + v^{5}}{{(u + v)}^{5}} = \frac{1 + t^{5}}{{(1 + t)}^{5}} = \frac{1 - t + t^{2} - t^{3} + t^{4}}{{(1 + t)}^{4}} = - \frac{1}{5}$ $6 - t + 11 t^{2} - t^{3} + 6 t^{4} = 0$ $(2 t^{2} - t + 2) (3 t^{2} + t + 3) = 0$ $\Rightarrow t^{2} - \frac{1}{2} t + 1 = 0 . . . (i)$ $\Rightarrow t^{2} + \frac{1}{3} t + 1 = 0 . . . (i i)$ $l e t \frac{u^{3} + v^{3}}{{(u + v)}^{3}} = \frac{1 + t^{3}}{{(1 + t)}^{3}} = \frac{1 - t + t^{2}}{{(1 + t)}^{2}} = \frac{1}{s}$ $s - s t + {s t}^{2} = 1 + 2 t + t^{2}$ $\Rightarrow t^{2} + \frac{2 + s}{1 - s} t + 1 = 0$ $w i t h \frac{2 + s}{1 - s} = - \frac{1}{2}$ $\Rightarrow s + 5 = 0 \Rightarrow s = - 5$ $w i t h \frac{2 + s}{1 - s} = \frac{1}{3}$ $\Rightarrow 4 s + 5 = 0 \Rightarrow s = - \frac{5}{4}$ $\Rightarrow \frac{u^{3} + v^{3}}{{(u + v)}^{3}} = \frac{1}{s} = - \frac{1}{5} o r - \frac{4}{5}$
	Commented by ajfour last updated on 08/Feb/19

	$T h a n k s S i r, y o u h a v e d o n e b e t t e r .$
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