Question

If a+b+c = 5, ab+bc+ca = 10, then prove that a^3+b^3+c^3 -3abc = -25

Answer 1

Answer:

Proved below

Step-by-step explanation:

a+b+c = 5

(a+b+c)^2 = 5^2=25

a^2+b^2+c^2 +2(ab+bc+ca)=25

a^2+b^2+c^2 + 2x10 = 25

a^2+b^2+c^2 = 25-20

a^2+b^2+c^2  = 5 -------eqn (1)

We know that

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

= (5){a^2+b^2+c^2 - (ab+bc+ca)}

 = (5)(5 - 10)     -------- from eqn (1)

= 5 (-5) = -25

Therefore a^3+b^3+c^3 -3abc = -25