Question

if a+b+c=5 and ab+bc+bc=10 then,prove that a^3+ b^3+c^3-3abc= -2​

Answer 1

Answer:

If a + b + c = 5 and ab + bc + ca = 10, then prove that a3+b3+c3−3abc=−25.

To prove, a3+b3+c3−3abc=−25,

Given,  

a+b+c=5,   ab+bc+ca=10

∵ (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

∴ (5)2=a2+b2+c2+2(10)

⇒ 25=a2+b2+c2+20

⇒ a2+b2+c2=25–20

⇒ a2+b2+c2=5

LHS=a3+b3+c3−3abc

=(a+b+c)(a2+b2+c2−ab−bc−ca)

=(5)[5−(ab+bc+ca)]

=5(5–10)=5(−5)=−25=RHS  

Hence proved.