Question

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

Answer 1

a+b+c = 5
Squaring on both sides, we get
a²+b²+c²+2(ab+bc+ca) = 5²
a²+b²+c²+2×10 = 25
a²+b²+c² = 25–20
a²+b²+c² = 5

a³+b³+c³–3abc = (a+b+c) (a²+b²+c²–(ab+bc+ca))
= 5×(5–10)
= 5×(–5)
= –25