Question

if a+b+c=15,and a ^2+b^2+c^2=83 find a^3+b^3+c^3=3abc

Answer 1

Answer:

Step-by-step explanation:

a+b+c=15--------------------------(1)

a²+b²+c²=83---------------------(2)  

square equation (1)

(a+b+c)² = 225  

⇒ a²+b²+c² + 2(ab+bc + ca)  =225

⇒ 83+ 2(ab+bc+ca)  =225

⇒ 2(ab+bc+ca) = 142

⇒ (ab+bc+ca) = 71

∴ a³+b³+c³-3abc

= {(a+b+c)(a²+b²+ c² - (ab+bc+ca)}

= 15 * (83-71)

= 15 * 12

= 180

Answer 2

==================================
⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐
==================================

Given, a+b+c = 15

Also, a²+b²+c² = 83


Now, a³+b³+c³-3abc

= (a+b+c) (a²+b²+c²-(ab+bc+ca))

= 15 ( 83-71 )

= 180.


THUS, THE ABOVE QUESTION GETS SOLVED.

==================================

*⃣ QUESTION IN YOUR MIND *⃣

⏩ How 71 came? What will be the procedure through which 71 came?

==================================

*⃣ ANSWER TO THIS QUESTION *⃣

⏩ (a+b+c)² = (15)²

=> a²+b²+c²+2(ab+bc+ca) = 225

=> 83 + 2(ab+bc+ca) = 225

=> ab + bc + ca = 142/2 or 71.


==================================

⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐

==================================