Question

solve this problem please​

Answer 1

hope the above picture will help you

Answer 2

Answer:

Step-by-step explanation:

a + b + c = 0

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

As given, a + b + c = 0

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

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

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

Now, divide $a^{3} + b^{3} + c^{3}$ and 3abc by abc

Divide abc on both sides.

So, it becomes,

$\frac{a^{3} + b^{3} + c^{3}}{abc}$=$\frac{3abc}{abc}$

$\frac{a^{2}}{bc} + \frac{b^{2} }{ca} +\frac{c^{2} }{ab} =3$

Therefore the value of $\frac{a^{2}}{bc} + \frac{b^{2} }{ca} +\frac{c^{2} }{ab}$ = 3

I HOPE THIS HELPS