Question

if a+b+c=15, a²+b²+c²=83, then find a³+b³+c³-3abc​

Answer 1

Step-by-step explanation:

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)

(15)^2=83+2(ab+bc+ca)

225–83=2(ab+bc+ca)

142/2=71=ab+bc+ca

a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

=(15)(83–71)

=15×12

=180 , Answer

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!