If a+b+c=5 and ab+bc+ca=10, then prove . . .

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

Doubt by Vikrant

Solution :

a+b+c=5 (Given)
ab+bc+ca=10 (Given)


We know,
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
(a+b+c)²=a²+b²+c²+2(ab+bc+ca)
(5)²=a²+b²+c²+2(10)
25=a²+b²+c²+20
25-20=a²+b²+c²
a²+b²+c²=5 — (1)

Also,
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
a³+b³+c³-3abc=(a+b+c)[a²+b²+c²-(ab+bc+ca)]
a³+b³+c³-3abc=(5)[5-(10)]
[Using equation (1)]
a³+b³+c³-3abc=(5)[-5]
a³+b³+c³-3abc=-25
Hence Proved.