Question

If a + b + c = 9 and a²+b²+c²= 35, find the value of a³+b³ +c³-3abc. (Solution is *108*)​

Answer 1

Step-by-step explanation:

Let,

a + b + c = 9 (1)

a² + b² + c² = 35 (2)

Squaring both sides

(a + b + c)² = (9)²

(a + b + c)² = a² + 2ab + 2ac + b² + 2bc + c²

a² + 2ab + 2ac + b² + 2bc + c² = 81

(a² + b² + c²) + 2ab + 2ac + 2bc = 81

Taking 2 as common

(a² + b² + c²) + 2(ab + ac + bc) = 81

From 2

(35) + 2(ab + ac + bc) = 81

2(ab + ac + bc) = 81 - 35

2(ab + ac + bc) = 46

ab + ac + bc = 46/2

ab + ac + bc = 23 (3)

Now

a³ + b³ + c³ - 3abc

a + b + c × a² + b² + c² - ab - ac - bc

(a + b + c)(a² + b² + c²) - (ab + ac + bc)

From 1,2 and 3

(9)(35 - 23)

(9)(12)

108

Answer 2

=

$a.2 + b.2 + c.2 + 2 (a b + bc + ca = 81 ) = 35 + 2(ab + bc + ca = 81 = ab + bc + ca = 23 (1) re call that (a.2 + b.2 + c.2 - ab - bc - (a) = 9(35 - 23) = 9(121) = 108$