Question

If a+b+c and ab+bc+ca=10 ,then prove that a cube+ b cube + c cube -3abc =-25

Answer 1

Answer:

If a+b+c = 5 and ab+bc+ca=10 , then a³ + b³ + c³ - 3abc  = -25

Step-by-step explanation:

If a+b+c and ab+bc+ca=10 ,then prove that a cube+ b cube + c cube -3abc =-25

a + b + c = 10  => a + b = 10-c

ab+bc+ca=10

a³ + b³ + c³ - 3abc

= (a + b)³ - 3ab(a+b) + c³ - 3abc

= (10 -c)³ - 3ab(10 -c) + c³ - 3abc

= 1000 - c³  -30c(10-c)  - 30ab + 3abc + c³ - 3abc

Cancelling c³ & 3abc  & putting 10-c = a + b

= 1000 - 30c(a + b) - 30ab

= 1000 - 30 (ac + bc + ab)

= 1000 - 30 (10)

= 1000 - 300

= 700

Let say a + b + c = 5  => a + b = 5-c

& ab+bc+ca=10

a³ + b³ + c³ - 3abc

= (a + b)³ - 3ab(a+b) + c³ - 3abc

= (5 -c)³ - 3ab(5 -c) + c³ - 3abc

= 125 - c³  -15c(5-c)  - 15ab + 3abc + c³ - 3abc

Cancelling c³ & 3abc  & putting 5-c = a + b

= 125 - 15c(a + b) - 15ab

= 125 - 15 (ac + bc + ab)

= 125 - 15 (10)

= 125 - 150

= -25