If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 – 3abc = – 25.
It is given in the question that,
a + b + c = 5
And, ab + bc + ca = 0
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
Putting the given values, we get:
(5)2 = a2 + b2 + c2 + 20
25 – 20 = a2 + b2 + c2
a2 + b2 + c2 = 5 (i)
Also, we know that
(a3 + b3 + c3 – 3abc) = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= (a + b + c) [(a2 + b2 + c2) – (ab + bc + ca)]
Now, putting the values we get:
= (5) × (5 – 10)
= 5 × (-5)
= - 25
Hence, it is proved that
(a3 + b3 + c3 – 3abc) = - 25