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