a-b-c-0-a-2-b-2-c-2-4-a-4-b-4-c-4-

Question Number 177331 by mr W last updated on 03/Oct/22

a + b + c = 0

a^{2} + b^{2} + c^{2} = 4

a^{4} + b^{4} + c^{4} = ?

Answered by a.lgnaoui last updated on 04/Oct/22

a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (a b + b c + a c) = 4

a b + b c + a c = - 2 (1) (

a^{4} + b^{4} + c^{4} = {(a^{2} + b^{2} + c^{2})}^{2} - 2 [{(a b)}^{2} + {(b c)}^{2} + {(a c)}^{2}]

a^{4} + b^{4} + c^{4} = 4^{2} - 2 [{(a b + b c + a c)}^{2} - (a b \times b c + a b \times a c + b c \times a b + b c \times a c + a c \times b c + a c \times a b) =

4^{2} - 2 [{(a b + b c + a c)}^{2} - 2 (a b \times b c + a b \times a c + b c a c)]

4^{2} - 2 [{(a b + b c + a c)}^{2} - 2 a b c (a + b + c)]

= 4^{2} - 2 [{(- 2)}^{2} - 0] = 8

d o n c a^{4} + b^{4} + c^{4} = 8

Commented by mr W last updated on 04/Oct/22

t h a n k s!

Commented by mr W last updated on 04/Oct/22

m a y i a s k y o u t o u s e a s m a l l e r f o n t

a n d u s e l i n e b r e a k ?

t h i s i s h o w y o u r p o s t s l o o k l i k e (d o w n)

a n d h o w o t h e r {p e o p l e}^{'} s p o s t s l o o k

l i k e (u p) . y o u r p o s t s a r e n o t g o o d f o r

r e a d i n g .

Commented by mr W last updated on 04/Oct/22

Commented by Ar Brandon last updated on 04/Oct/22

Commented by Ar Brandon last updated on 04/Oct/22

Commented by mr W last updated on 04/Oct/22

t h a n k s s i r!

f o n t s i z e i s j u s t o n e p r o b l e m . t h e

o t h e r p r o b l e m i s l i n e b r e a k .

Commented by Ar Brandon last updated on 04/Oct/22

It's true Sir. But we may also use zoom in, zoom out, fit width.

Commented by mr W last updated on 04/Oct/22

c e r t a i n l y! w e c a n l i v e w i t h i t! i p e r f e r

s u c h p o s t s w h i c h c a n b e r e a d w i t h o u t

s c r o l l i n g c o n s t a n t l y f r o m l e f t t o r i g h t

a n d f r o m r i g h t t o l e f t . w h e n u s i n g

“ f i t {w i d t h}^{″}, t e x t w i t h o u t l i n e b r e a k

a p p e a r s i n t i n y f o n t .

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