Math, asked by subbu2000, 1 year ago

If a+b+c=15 and ab+bc+ca=74. Find (a+b+c).

Answers

Answered by pal69
1

Answer:

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

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