Question

a+b+c= 5 ab+bc+ca=10 prove that cube of a+cube of b+cube of c-3abc=25

Answer 1

$<b>Hey mate!$

Given:

We know ,

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

now ,

a + b + c = 5

ab + bc + ca = 10

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

=> (5)² -2×10 = a² + b² + c²

=>a² + b² + c² =5

Now,

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

=> 5 (5 - 10)

=> 5 × (-5)

=> -25

Hence, proved...



Thanks for the question!