Question

Os. No: 7(b)
Find a2+b2+c2 if a+b+c= 20 and ab+bc+ca=71.​

Answer 1

Answer:

(a + b + c)2 = a2 +b2 + c2 + 2ab + 2bc + 2ca. ⇒ (a + b + c)2 = 83 + 2(ab + bc + ca). ⇒ (a + b + c)2 = 83 + 2 × 71. ⇒ (a + b + c)2 = 83 + 142.

Answer 2

Answer:

there is one identity

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

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

=(20)^2-2*71

=400-142

=258