a + b + C is equal to 12, a square + b square + c square is equal to 90 find the value of a cube plus b cube plus c cube minus a b c
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Answered by
13
(a+b+c)=12
a^2+b^2+c^2=90
A^3+b^3+c^3-abc
I think if we can use a=9 b=3 and =0
as 9+3+0=12&81+9+0=90
so 9^3+3^3+0^3-3★9★0=756
a^2+b^2+c^2=90
A^3+b^3+c^3-abc
I think if we can use a=9 b=3 and =0
as 9+3+0=12&81+9+0=90
so 9^3+3^3+0^3-3★9★0=756
Answered by
1
Answer:
Value of a³ + b³ + c³ - 3abc is 756.
Step-by-step explanation:
We are given, a + b + c = 12 and a² + b² + c² = 90
To find: a³ + b³ + c³ - 3abc
First we use following identity to find value of ab + cb + ac,
( a + b + c )² = a² + b² + c² + 2ab + 2cb + 2ac
( 12 )² = 90 + 2ab + 2cb + 2ac
2 (ab + cb + ac) = 144-90
ab + cb + ac = 54/2
ab + cb + ac = 27
Now using identity,
a³ + b³ + c³ - 3abc = ( a + b + c)( a² + b² + c² - ab - cb - ac )
= ( 12 ) ( 90 - (ab + cb + ac))
= ( 12 ) ( 90 - (27))
= ( 12 ) ( 63 )
= 756
Therefore, Value of a³ + b³ + c³ - 3abc is 756.
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