Question

plz solve this
a+b+c =5 and ab+ bc+ac=10
then prove
a3 + b3+c3 _ 3abc = _25
3is the exponent here not the co ficient .
I need your help.

Answer 1

$<b>$

here

identity we need to prove :-

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

Given:

(a+b+c) and ab+bc+ca,

Find:-

a2+b2+c2.





(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

​(a+b+c)² = a²+b²+c2+2(ab+bc+ca)

Now,

substitute the values.

(5)² = a2+b2+c2+2(10)

25 = a2+b2+c2+20

25-20 = a2+b2+c2

a2+b2+c2 = 5

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

-25 = 5(5-10)

-25= 5×(-5)

-25 = -25

LHS=RHS

Hence Proved