Question

If a+b+c=3 and ab+ bc+ca=10 then find the value of a^3+b^3+c^3-3abc

Answer 1

Answer:

Step-by-step explanation:

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

now ,

a + b + c = 5

ab + bc + ca = 10

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

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5

hence ,

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

=( 5)( 5 - 10) = 5 × (-5) = -25

hence proved

Answer 2

Answer:

We know ,

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

now ,

a + b + c = 5

ab + bc + ca = 10

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

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5

hence ,

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

=( 5)( 5 - 10) = 5 × (-5) = -25