Question

If a + b + c = 15 and a2 + b2 + c2 = 83, find the value of a3 + b3 + c3 3abc.

Answer 1

Answer:

180

Step-by-step explanation:

a+b+c=15

on squaring both the sides

(a+b+c)² =15²

a²+b²+c²+2ab+2bc+2ca=225

83+2ab+2bc+2ca=225 (a²+b²+c²=83)

2ab+2bc+2ca=225-83

2ab+2bc+2ca=142

2(ab+bc+ca)=142

ab+bc+ca=142/2=71

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

a³+b³+c³-3abc=(15){83-(ab+bc+ca)}

a³+b³+c³-3abc=(15){83-71)

a³+b³+c³-3abc=(15)(12)

a³+b³+c³-3abc=180

Answer 2

Step-by-step explanation:

Given: a + b + c = 15, a² + b² + c² = 83

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

⇒ 15² = 83 + 2(ab + bc + ca)

⇒ 225 - 83 = 2(ab + bc + ca)

⇒ 142 = 2(ab + bc + ca)

⇒ ab + bc + ca = 71

Now,

a³ + b³ + c³ - 3abc:

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

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

= (15)(83 - 71)

= 180.

Hence, the value of a³+b³+c³-3abc is 180

Hope it helps!