Question

if a+b+c = 5 and ab+bc+ca = 10
proove a³+b³+c³-3abc = -25

Answer 1

Answer:

a³ + b³ + c³ - 3abc = -25

Step-by-step explanation:

Given, a + b + c = 5

On squaring both sides

(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

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

                                                       = 5× (5 - 10)

                                                       = 5×(-5) = -25

Answer 2

We are given :-

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

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

=> 5² = a² + b² + c² + 2(10)

=> a² + b² + c² = 5

Now, we know that,

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

=> a³ + b³ + c³ - 3abc = (5)[5 - 10] = -25

Q.E.D.

Hope it'll help you.....