Question

if a+b+c=0 then show that a2(b+c)+b2(c+a)+c2(a+b)+3abc=0​

Answer 1

$a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0$, proved.

Step-by-step explanation:

We have,

a + b + c = 0

Show that, $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0$.

∵ a + b + c = 0

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

L.H.S. = $a^2$(b + c)+ $b^2$(c + a) + $c^2$(a + b) + 3abc

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

= $a^2$(- a)+ $b^2$(- b) + $c^2$(- c) + 3abc

= $-a^{3} -b^{3}-c^{3}$ + 3abc

= - ($a^{3}+b^{3}+c^{3}$ - 3abc)

We know that,

$a^{3}+b^{3}+c^{3}$ - 3abc = (a + b + c)($a^{2} +b^{2} +c^{2} -ab-bc-ca)$

= - ($a^{3}+b^{3}+c^{3}$ - 3abc)= (0)($a^{2} +b^{2} +c^{2} -ab-bc-ca)$

= 0

= R.H.S., Proved

Thus, $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0$, proved.

Answer 2

Answer:

It is proved that, $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0.$

Step-by-step explanation:

We have,

$a+b+c=0$

Show that, $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0$

$\because a+b+c=0$

$\Rightarrow a+b=-c,b+c=-a$

$L.H.S=a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc$

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

$=a^2(-a)+b^2(-b)+c^2(-c)+3abc$

$=-a^3-b^3-c^3+3abc$

$=-(a^3+b^3+c^3-3abc)$

We know that,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=-(a^3+b^3+c^3-3abx)=(0) (a^2+b^2+c^2-ab-bc-ca)$

$=0$

$=R.H.S.$ Proved

Thus, $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0,$ proved.

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