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