Math, asked by shaams, 1 year ago

if a+b+c=15 and a^2+b^2+c^2=83 find the value of a^3+b^3+c^3-3abc


nemo29: mark as brainliest

Answers

Answered by nemo29
2
HERE'S YOUR ANSWER......✌️✌️
Attachments:

shaams: thank you
nemo29: welcome
Answered by Salmonpanna2022
1

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!

Similar questions