Hey mates!!!
Plz help me to solve this question....
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Anonymous:
Hey Sanskriti , can I solve it ?
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Answered by
0
it is given .
asqu/bc+b2/ac+c2/ab
we can write this as
a3+b3+c3/abc
a+b+C=0 so a3+b3+c3=3abc
putting value of a3+b3+c3
3abc/abc
=3
hence proved
asqu/bc+b2/ac+c2/ab
we can write this as
a3+b3+c3/abc
a+b+C=0 so a3+b3+c3=3abc
putting value of a3+b3+c3
3abc/abc
=3
hence proved
how you write a3b3c3/abc please tell
thanks
by taking LCM
^_^
Answered by
3
Hey Mate !
Here is your solution :
Given,
=> a + b + c = 0
=> a + b = -c --------- ( 1 )
Cubing both sides,
=> ( a + b )³ = ( -c )³
=> a³ + b³ + 3ab( a + b ) = -c³
Substitute the value of ( 1 ),
=> a³ + b³ + 3ab( -c ) = -c³
=> a³ + b³ - 3abc = -c³
•°• a³ + b³ + c³ = 3abc ------- ( 3 )
Now,
=> ( a²/bc ) + ( b²/ac ) + ( c²/ab ) = 3
=> ( a³ + b³ + c³ ) / abc = 3
Substitute the value of ( 3 ),
=> 3abc / abc = 3
•°• 3 = 3
☆ Proved ☆
=============================
Hope it helps !! ^_^
Here is your solution :
Given,
=> a + b + c = 0
=> a + b = -c --------- ( 1 )
Cubing both sides,
=> ( a + b )³ = ( -c )³
=> a³ + b³ + 3ab( a + b ) = -c³
Substitute the value of ( 1 ),
=> a³ + b³ + 3ab( -c ) = -c³
=> a³ + b³ - 3abc = -c³
•°• a³ + b³ + c³ = 3abc ------- ( 3 )
Now,
=> ( a²/bc ) + ( b²/ac ) + ( c²/ab ) = 3
=> ( a³ + b³ + c³ ) / abc = 3
Substitute the value of ( 3 ),
=> 3abc / abc = 3
•°• 3 = 3
☆ Proved ☆
=============================
Hope it helps !! ^_^
I dont know that ur here..Sorry!
no problem bro
^_^
^_^
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why did the modeartors delete his account
Don't know sis
moderators r very rude
Don't talk here
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