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Question :

If a+ b + c= 15 and a²+ b²+c²=83, find the value of a³+ b³+c³-3abc.

Solution :

According to the question,

a + b + c = 15

a² + b² + c² = 83

So by squaring both sides :

→ (a + b + c)² = (15)²

As we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

So it would be helpful for getting ab + bc + ca

→ a² + b² + c² + 2(ab + bc + ca) = 225

By putting the value of a² + b² + c² we get

→ 83 + 2(ab + bc + ca) = 225

→ 2(ab + bc + ca) = 225 - 83

→ 2(ab + bc + ca) = 142

By transposing 2 on the R.H.S we get

→ ab + bc + ca = 142/2

→ ab + bc + ca = 71

Now to get a³ + b³ + c³- 3abc

a³+ b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Now - ab - bc - ca can be taken as -1(ab + bc + ca)

a³+ b³ + c³ - 3abc = (a + b + c)(a² + b² + c² -1(ab + bc + ca))

By putting the values we get

a³+ b³ + c³ - 3abc = 15(83 - 71)

a³+ b³ + c³ - 3abc = 15 × 12

a³+ b³ + c³ - 3abc = 180

So the value is 180

Explanation:

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