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

Solution:

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

We have to prove that a³ + b³ + c³ - 3abc = - 25.

Using the algebraic identity,

x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)

x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - (xy + yz + zx))

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

= 5(a² + b² + c² - (10))

To find a² + b² + c², by using the algebraic identity,

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

(5)² = a² + b² + c² + 2(10)

25 = a² + b² + c² + 20

a² + b² + c² = 25 - 20

a² + b² + c² = 5

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

= 5(5 - 10)

= 5(-5)

= - 25

Therefore, a³ + b³ + c³ - 3abc = - 25

✦ Try This: Find the remainder when f(x) = 4x³ - 12x² +14x - 3 is divided by g(x) = (2x-1).

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2

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NCERT Exemplar Class 9 Maths Exercise 2.4 Problem 8

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

Summary:

If a + b + c = 5 and ab + bc + ca = 10, using the algebraic identity it is proven that a³ + b³ + c³ - 3abc = -25

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