If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ – 3abc = – 25.

Question

If a + b + c = 5 and ab + bc + ca = 10, prove that a 3 + b 3 + c 3 – 3abc = – 25.

Philoid · Accepted Answer

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

If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 – 3abc = – 25.