Question

If a+b+c=6, abc=6, ab+bc+ac=11, find a^3 + b^3 + c^3

Answer 1

Answer:

a³+b³+c³=36

Step-by-step explanation:

Given:-

a+b+c=6,abc=6,ab+bc+ca=11

To Find:-

a³+b³+c³

Formulas Applied:-

1. (a+b+c)²=a²+b²+c²+2(ab+bc+ca)
2. a³+b³+c³-3abc=(a+b+c)[a²+b²+c²-(ab+bc+ca)]

Solution:-

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

(6)²=a²+b²+c²+2(11)

a²+b²+c²=36-22

a²+b²+c²=14

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

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

a³+b³+c³=(6)[(14)-11]+3(6)

a³+b³+c³=(6)(3)+18

a³+b³+c³=18+18

a³+b³+c³=36

Answer 2

Answer:

a³+b³+c³=36

Step-by-step explanation:

Given:-

a+b+c=6,abc=6,ab+bc+ca=11

To Find:-

a³+b³+c³

Formulas Applied:-

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

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

Solution:-

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

(6)²=a²+b²+c²+2(11)

a²+b²+c²=36-22

a²+b²+c²=14

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

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

a³+b³+c³=(6)[(14)-11]+3(6)

a³+b³+c³=(6)(3)+18

a³+b³+c³=18+18

a³+b³+c³=36h