Question

if
a+b+c= 12 , and 
{a}^{2} + {b}^{2} + {c }^{2} = 90a​2​​+b​2​​+c​2​​=90 
find the value of 
{a}^{3} + {b}^{3} + {c}^{3} - 3abca​3​​+b​3​​+c​3​​−3abc 

Answer 1

its
=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
=12×(90-ab-(12-a-b)(a+b)
now I think you can solve further by opening it

Answer 2

Answer:

756

Step-by-step explanation:

A+b+c= 12 and a 2 + b 2 +c 2 = 90, find the value of a 3 +b 3 +c 3 -3abc  

Consider the formula - a³+b³+c³- 3abc = (a+b+c) (a²+b²+c²-(ab+bc+ca))

We have to find ab+bc+ca

given a+b+c = 12

Squaring on both sides we get,

(a+b+c)² = 12²

a²+b²+c² + 2(ab+bc+ca) = 144

2 (ab+bc+ca) = 54

ab + bc + ca = 27

Now, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-(ab+bc+ca))

Putting the values we get

12 (90 - 27)

12 x 63

756