Question

the value of a cube plus b cube plus c cube minus 3 ABC if a + b + c is equals to 15 and a b + BC + CA is equal to 74

Answer 1

Given ,
${x}^{3} + {y}^{3} + {z}^{3} - 3xyz$
$= (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$

Answer 2

Given: a + b + c = 15 and ab + bc + ca = 74

To find: a^3 + b^3 + c^3

Identities to be used:

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

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

Answer:

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

Since we don't have the value for a^2 + b^2 + c^2, we will be using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

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

15^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

225 = a^2 + b^2 + c^2 + 2*74

225 = a^2 + b^2 + c^2 + 148

a^2 + b^2 + c^2 + 148 = 225

a^2 + b^2 + c^2 = 225 - 148

a^2 + b^2 + c^2 = 77

Now we know the value of a^2 + b^2 + c^2, so we will go back to using the first identity to find the value of a^3 + b^3 + c^3 - 3abc

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 * (a^2 + b^2 + c^2 - (ab + bc + ca))

a^3 + b^3 + c^3 - 3abc = 15*(77 - 74)

a^3 + b^3 + c^3 - 3abc = 15*3

a^3 + b^3 + c^3 - 3abc = 45

Hope it helps :)