Question

If a+b+c=15 and asquare+bsquare+csquare=83, find the value of acube+bcube+ccube-3abc

Answer 1

Answer:

Step-by-step explanation:

a + b + c = 15

(a + b + c)^2 = 15^2

a^2 + b^2 + c^2 + 2(ab + bc +ca) =225

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

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

2(ab + bc +ca) = 142

ab + bc +ca = 142/2

ab + bc +ca = 71

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

a^3 + b^3 + c^3 - 3abc = (15)(83 + 71)

a^3 + b^3 + c^3 - 3abc = (15)(154)

a^3 + b^3 + c^3 - 3abc = 2310

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!