Question

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

Answer 1

Answer is 180

Step-by-step explanation:

(a+b+c)² = a²+b²+c²+ 2(ab+bc+ac)

=>(15)² = 83+2(ab+bc+ac)

=>(ab+bc+ac) = (225-83)/2

(ab+bc+ac) = 71

now

a³+b³+c³-3abc

= (a+b+c) (a²+b²+c²-ab-bc-ac)

= 15 X (83-71)

= 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!