Question

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

Answer 1

Given: a + b + c = 5 and ab + bc + ca = 10

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)] 

= 5{a² + b² + c² – (ab + bc + ca)}

= 5(a² + b² + c² – 10)

Given: a + b + c = 5

Now, squaring both sides, get: (a + b + c)² = 5²

a² + b² + c² + 2(ab + bc + ca) = 25

a² + b² + c² + 2 × 10 = 25

a² + b² + c² = 25 – 20

= 5

Now, a³ + b³ + c³ – 3abc = 5(a² + b² + c² – 10)

= 5 × (5 – 10)

= 5 × (–5)

= –25

Hence proved.