Question

$if \: a \: + b \: + c \: = \: 15 \: and \: {a}^{2} + {b}^{2} + {c}^{2} = 83 \\ find \: the \: value \: of {a}^{3} + {b}^{3} + {c }^{3} - 3abc$

Answer 1

Here is solution of your question
a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)
a²+b²+c²=83
(a+b+c)²-2ab-2bc-2ca=83
15²-2(ab+bc+ca)=83
-2(ab+bc+ca)=83-225
-2(ab+bc+ca) =-142
ab+bc+ca=-142/-2
=71
then
15(83-(71)
15×12
180Ans

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!