Question

find a3+b3+c3-3abc when a+b+c=15 ab+bc+ca=74

Answer 1

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Answer 2

Answer :

a³ + b³ + c³- 3abc = 45

Step-by-step explanation :

Given,

a + b + c = 15

ab + bc + ca = 74

We need to find the value of a³ + b³ + c³- 3abc

We know,

a³ + b³ + c³- 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )

Now we need to find the value of  a² + b² + c²

We also know,

( a + b + c )² = a² + b² + c² + 2 ( ab + bc + ca )

Putting the value of a + b + c = 15 and ab + bc + ca = 74

( 15 )² = a² + b² + c² + 2 ( 74 )

225 = a² + b² + c² + 148

a² + b² + c² = 225 - 148

a² + b² + c² = 77

Now,

Substituting value in the formula for a³ + b³ + c³- 3abc

a³ + b³ + c³- 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )

a³ + b³ + c³- 3abc = ( a + b + c ) ( a² + b² + c² - ( ab + bc + ca ) )

a³ + b³ + c³- 3abc = ( 15 ) ( 77 - ( 74 ) )

a³ + b³ + c³- 3abc =  ( 15 ) ( 3 )

a³ + b³ + c³- 3abc = 45

Hence,

a³ + b³ + c³- 3abc = 45