Question

If a+b+c=10 and ab+bc+ca=12 then find the value of a³ +b³+c³-3abc.​

Answer 1

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)...............(1)

we have,

a+b+c=10.........(2),

ab+bc+ca=12.........(3)

we have to find the value of a²+b²+c²

so,

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

10²=a²+b²+c²+2(12)

a²+b²+c²= 100-24 = 76.........(4)

put (2),(3),(4) in equation (1)

we, get

a³+b³+c³-3abc= 10[76-(12)]

= 10(64)

= 640

Answer 2

Answer:

640

Step-by-step explanation:

a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca).....(1)

a+b+c=10

ab+bc+ca=12

to find a2+b2+c2:

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

10^2-2(12)=a2+b2+c2

a2+b2+c2=100-24

a2+b2+c2=76

substituting in (1)

a3+b3+c3-3abc=(10)(76-(ab+bc+ca))

a3+b3+c3-3abc=(10)(76-12)

a3+b3+c3-3abc=(10)(64)

a3+b3+c3-3abc=640

:):):):):):)