Determine the formula for a2+b2+c2
STEP 1 : Expanding (a+b+c)2
We know that (a+b+c)2 can be expressed as
(a+b+c)2=(a+b+c)(a+b+c)
(a+b+c)2=a2+ab+ac+ba+b2+bc+ca+cb+c2
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
STEP 2 : Transposing 2ab+2bc+2ca on L.H.S.
Transposing 2ab+2bc+2ca on L.H.S. we get
(a+b+c)2-2ab-2bc-2ca=a2+b2+c2
⇒a2+b2+c2=(a+b+c)2-2ab-2bc-2ca
On further simplification we get
⇒a2+b2+c2=(a+b+c)2-2(ab+bc+ca)
Hence, the formula for a2+b2+c2 is a2+b2+c2=(a+b+c)2-2(ab+bc+ca).
If a + b + c = 2s, then prove the following identities
(a) s2 + (s − a)2 + (s − b)2 + (s − c)2 = a2 + b2 + c2
(b) a2 + b2 − c2 + 2ab = 4s (s − c)
(c) c2 + a2 − b2 + 2ca = 4s (s − b)
(d) a2 − b2 − c2 + 2ab = 4(s − b) (s − c)
(e) (2bc + a2 − b2 − c2) (2bc − a2 + b2 + c2) = 16s (s − a) (s − b) (s − c)
(f)