if (xyz^(−) )^2 = (x+y+z)^5 then find: x^3 +y^3 +z^3 −∣(x+y+z)+(x^2 +y^2 +z^2 )∣


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Question Number 147187 by mathdanisur last updated on 18/Jul/21

$i f {(\overset{―}{x y z})}^{2} = {(x + y + z)}^{5} t h e n f i n d :$ $x^{3} + y^{3} + z^{3} - ∣ (x + y + z) + (x^{2} + y^{2} + z^{2}) ∣$
Commented by Rasheed.Sindhi last updated on 18/Jul/21

$I s \overset{―}{x y z} a d i g i t a l r e p r e s e n t a t i o n o f$ $d e c i m a l n u m b e r ?$
Commented by mathdanisur last updated on 18/Jul/21

$Y e s S e r$
	Answered by Rasheed.Sindhi last updated on 19/Jul/21

	${(\overset{―}{x y z})}^{2} = {(x + y + z)}^{5}$ $▸ \overset{―}{x y z} i s 3 - d i g i t n a t u r a l n u m b e r$ $▸ {(\overset{―}{x y z})}^{2} i s p e r f e c t 5 t h p o w e r$ $\Rightarrow \overset{―}{x y z} i s p e r f e c t 5 t h p o w e r$ $T h e o n l y t h r e e d i g i t n u m b e r w h i c h$ $i s a l s o 5 t h p o w e r i s 243 (= 3^{5})$ $[2^{5} = 32 2 - d i g i t n u m b e r \times$ $3^{5} = 243 3 - d i g i t n u m b e r ✓$ $4^{5} = 1024 4 - d i g i t n u m b e r \times]$ $P o s s i b l e c a n d i d a t e i s o n l y 243 a n d$ $i t a l s o f u l f i l l s t h e g i v e n c o n d i t i o n :$ ${(243)}^{2} \overset{?}{=} {(2 + 4 + 3)}^{5}$ ${(3^{5})}^{2} \overset{?}{=} 9^{5}$ ${(3^{5})}^{2} \overset{?}{=} {(3^{2})}^{5}$ $3^{10} = 3^{10}$ $∴ 243 i s o n l y s u c c e s s f u l c a n d i d a t e .$ $∴ x = 2, y = 4, z = 3$ $x^{3} + y^{3} + z^{3} - ∣ (x + y + z) + (x^{2} + y^{2} + z^{2}) ∣$ $= 2^{3} + 4^{3} + 3^{3} - ∣ (2 + 4 + 3) + (2^{2} + 3^{2} + 4^{2} ∣$ $= 8 + 64 + 27 - ∣ 9 + 29 ∣$ $= 61$
	Commented by mathdanisur last updated on 19/Jul/21

	$c o o l S e r t h a n k s$
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