Question

If a + b + c = 4 and a² + b² + c² = 16, find the value of a³ + b³ + c³- 3abc.​

Answer 1

Answer:     112

STEP BY STEP

Given =>

a + b + c = 16

a² + b² + c² = 90

To find =>

a³ + b³ + c³ - 3abc

We know,

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

But, we have to find out a³ + b³ + c³ - 3abc and in above formula we know the value of a + b + c and a² + b² + c² but, not the value of ab + bc + ca

So, let's find out value of - ab - bc - ca by this,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Now, let's put the values,

(16)² = 90 + 2(ab + bc + ca)

256 = 90 + 2(ab + bc + ca)

256 - 90 = 2(ab + bc + ca)

166 = 2(ab + bc + ca)

ab + bc + ca = 166/2

ab + bc + ca = 83

Now, we know all required values to find out a³ + b³ + c³ - 3abc let's put them in formula

a³ + b³ + c³ - 3abc = (16) [90 - (83)]

a³ + b³ + c³ - 3abc = 16 * 7

a³ + b³ + c³ - 3abc = 112

The value of a³ + b³ + c³ - 3abc is 112.

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