Question

Q 45.
If a+b+c = 1 and ab+bc+ca = 17, then what is value of a+b3+c3-3abc?
(a) 121
ho
(b) 168
(C) 0 300
(d) 6 770​

Answer 1

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

Step-by-step explanation:

please mark me brainliest and follow me

Answer 2

answer is 18 please mark me as a brainliest.