Question

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

Answer 1

$a + b + c = 15 \: \: \: \: \: and \: \: \: \: {a}^{2} + {b}^{2} + {c}^{2} = 83 \\ (a + b + c) ^{2} = {15}^{2} \\ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 225 \\ 2(ab + bc + ca) = 225 - 83 \\ 2(ab + bc + ca) = 142 \\ ab + bc + ca = 71 \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)[ a^{2} + b ^{2} + c ^{2} - (ab + bc + ca) ] \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 15(83 - 71) \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc= 15 \times 12 \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc= 180$

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!