Question

Plz solve this question .​

Answer 1

Step-by-step explanation:

it is given that ,

a+b+c=0             (1)

and we need to show that

$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+3abc=0$

from equation number (1),we get

a+b+c=0

b+c = -a     (2)

a+b+c=0

c+a = -b     (3)

a+b+c=0

a+b =  -c    (4)

now,putting the values of equation (2),(3)and (4) in given equation .

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

so,

that is a little try from me to solve this equation hope you will understand.

Answer 2

Answer:

my name is swee.ty

Step-by-step explanation:

Acceleration is a vector quantity as it has both magnitude and direction. It is also the second derivative of position with respect to time or it is the first derivative of velocity with respect to time.