Math, asked by bbJkao, 11 months ago

यदि a + b + c = 8 एवं a² + b² + c² = 24 हो तो a³ + b³ + c³ - 3abc का मान क्या होगा​

Answers

Answered by Swarnimkumar22
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 Darvince
7

Answer:

Given that - a + b + c = 8, a² + b² + c² = 24 and a³ + b³ + c³ - 3abc

\longrightarrow ( a + b + c)² = 64

\longrightarrow a² + b² + c² + 2(ab + bc + ca) = 64

\longrightarrow (Place the value of a² + b² + c²)

\longrightarrow 24 + 2(ab + bc + ca) = 64

\longrightarrow Get 24 on the other side

\longrightarrow 2(ab + bc + ca) = 64 - 24

\longrightarrow ab + bc + ca = 40/2

\longrightarrow ab + bc + ca = 20

_________________________

Value of a³ + b³ + c³ - 3abc -

\longrightarrow (a + b + c)(a² + b² + c² - ab - bc - ca)

  • a + b + c = 8
  • a² + b² + c² = 24
  • ab + bc + ca = 20

Place them,

\longrightarrow 8(24 - 20)

\longrightarrow 8(4)

\longrightarrow 32

.°. a³ + b³ + c³ - 3abc = 32

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