Question

if a+b+c=0 than prove that a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0​

Answer 1

Answer:

a²(b + c) + b² (c + a) + c²(a + b) + 3abc = 0,

proved.

Step-by-step explanation:

We have,

a+b+c=0

Show that, a²(b + c) + b² (c + a) + c² (a + b) + 3abc = 0.

a+b+c=0

⇒ a + b = - c, b + c = -a and c + a = - b

L.H.S. = a² (b + c)+ b² (c + a) + c² (a + b) + 3abc

Put a + b = - c, b + c = - a and c + a = - b, we get

= a² (− a)+ b² (- b) + c² (- c) + 3abc

= −a³ − b³ − c³ + 3abc =

= - (a³ + b³ + c³ - 3abc)

We know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² +6² + c² - ab — bc - ca)

- (a³ + b³ + c³ - 3abc)= (0)(a² +6² + c²ab-bc- ca)

=

= R.H.S., Proved

Thus, a²(b + c) + b² (c + a) + c²(a + b) + 3abc = 0, proved.