यदि a + b + c = 8 एवं a² + b² + c² = 24 हो तो a³ + b³ + c³ - 3abc का मान क्या होगा
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Answered by
5
solution :-
a + b + c = 8
=> ( a + b + c)² = 64
=> a² + b² + c² + 2(ab + bc + ca) = 64
=> 24 + 2 (ab + bc + ca) = 64
=> 2(ab + bc + ca) = 64 - 24
=> ab + bc + ca = 40/2 = 20
पुनः
a³ + b³ + c³ - 3abc
= ( a+ b+ c) (a² + b² + c² - ab -bc -ca)
= 8(24-20 = 8×4 = 32
Answered by
7
Answer:
Given that - a + b + c = 8, a² + b² + c² = 24 and a³ + b³ + c³ - 3abc
( a + b + c)² = 64
a² + b² + c² + 2(ab + bc + ca) = 64
(Place the value of a² + b² + c²)
24 + 2(ab + bc + ca) = 64
Get 24 on the other side
2(ab + bc + ca) = 64 - 24
ab + bc + ca = 40/2
ab + bc + ca = 20
_________________________
† Value of a³ + b³ + c³ - 3abc -
(a + b + c)(a² + b² + c² - ab - bc - ca)
- a + b + c = 8
- a² + b² + c² = 24
- ab + bc + ca = 20
Place them,
8(24 - 20)
8(4)
32
.°. a³ + b³ + c³ - 3abc = 32
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