Question

a+b+c=38 and a²+b²+c²=7202 and find the value of abc

Answer 1

Answer:

I derived the following two relations:

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

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

Using the above two equations (1) and (2), we can find the answer to abc. We are given,

a+b+c=6, a²+b²+c²=14 and a³+b³+c³=36……………………………..…………………(A)

From eq.(2),

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

= 6² - 14 = 36–14=22 [Using eq.(A)]

Or, ab+bc+ca = 11………………………………………………………………………………(B)

We now substitute the numerical values given in equations (A) and (B) into eq.(1) and obtain,

36 - 3abc = 6³ - 3 x 11 x 6 = 216 - 198 = 18

Or, -3abc = 18–36=-18 This yields

abc = -18/-3 = 6 (Proved)