Question

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

Answer 1

here is your Answer brother!
$(a + b + c) {}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) \\ now \: put \: the \: value \\ \\ (15) {}^{2} = 83 + 2(ab + bc + ca) \\ 2(ab + bc + ca) = 225 - 83 \\ ab + bc + ca = \frac{142}{2} = 71 \\ \\ now \: \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca)) \\ = now \: putting \: value \\ 15 \times (83 - 71) \\ = 15 \times 12 \\ = 180 \: ans \\ hope \: its \: helpful \: for \: u$

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!