Polynomials with difficulty......

$a+b+c$ = $0$

$a+b$ = $-c$

$a^{2}+b^{2}+2ab$ = $c^2$

$a^{2}+b^{2}-c^{2}$ = $-2ab$

$({a^{2}+b^{2}-c^{2})}^2$ = $4a^{2}b^{2}$

$a^{4}+b^{4}+c^{4}+2a^{2}b^{2}-2b^{2}c^{2}-2c^{2}a^{2}$ = $4a^{2}b^{2}$

$a^{4}+b^{4}+c^{4}$ = $2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}$

$a^{4}+b^{4}+c^{4}$ = $2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$

Hence,the result of the given expression is 2.

$a+b+c=0$ Squaring both sides we get :

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2ab+2bc+2ac=0$

Rearranging we get :

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=-2(ab+bc+ac)$ again squaring both sides we get :

${ a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }+2({ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 })=4({ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 }+2abc(a+b+c))$

Using $a+b+c=0$ and rearranging some terms we get :

${ a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }=2({ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 })$

Finally $\boxed { \frac { { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 } }{ { a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 } } =2 }$

People I have been getting many notifications on clarification of the case if we take 0. I would like to tell you if we take a,b and c as 0, the situation comes out to be 0/0. Remember 0/0 is not defined as any no. divided by 0 is undefined so 0/0 can be any answer so in these questions we don't take it as 0/0 as answer. By the way 0/0 can also be the answer too. I hope I have cleared your doubt but do comment if it still persists and the way Anik Mandal solved is just the way you do it. Do share your different method of solving it:)

its simple for this problem does'nt need to think big take 3 values satisfying a+b+c=0 as a=2 b=-1 c=-1 thats it we get solution easily

a + b + c = 0

(a + b + c)^2 = 0

a^2 + b^2 + c^2 + 2(ab + bc + ac) = 0

a^2 + b^2 + c^2 = -2 (ab + bc + ac)

[a^2 + b^2 + c^2]^2 = [-2 (ab + bc + ac)]^2

a^4 + b^4 + c^4 + 2[a^2.b^2 + b^2.c^2 + a^2.c^2] = 4[a^2.b^2 + b^2.c^2 + a^2.c^2 + 2abc (a + b + c ) ]

But a + b + c = 0 a^4 + b^4 + c^4 + 2[a^2.b^2 + b^2.c^2 + a^2.c^2] = 4[a^2.b^2 + b^2.c^2 + a^2.c^2]

a^4 + b^4 + c^4 = 2[a^2.b^2 + b^2.c^2 + a^2.c^2]

Hence, a^4 + b^4 + c^4 / [a^2.b^2 + b^2.c^2 + a^2.c^2] = 2

Put any three numbers a,b,c such that a+b+c = 0, and you'll get same answer in each case!! That's because its actually an identity in a,b,c for all real a+b+c = 0. This method is meant only to get quick answers and is not actual solution to the problem.

a+b+c = 0 --> a+b = -c -->(a+b)^4 = (-c)^4 -->a^4 + (4a^3)b + (6a^2)(b^2) + 4ab^3 + b^4 = c^4

a^4 + b^4 + c^4 --> a^4 + b^4 + a^4 + (4a^3)b + (6a^2)(b^2) + 4ab^3 + b^4 = 2a^4 + (4a^3)b + (6a^2)(b^2) + 4ab^3 + 2b^4

(a+b)^2 = (-c)^2 -->(a^2 + 2ab + b^2) = c^2 -->(a^2)(b^2) + (b^2)(c^2) + (c^2)(a^2) --> (a^2)(b^2) + (a^2 + 2ab + b^2)b^2 + a^2(a^2 + 2ab + b^2) = a^4 + (2a^3)b + (3a^2)(b^2) + 2ab^3 + b^4

So, there fore... (2a^4 + (4a^3)b + (6a^2)(b^2) + 4ab^3 + 2b^4)/(a^4 + (2a^3)b + (3a^2)(b^2) + 2ab^3 + b^4)

= 2 (Q.E.D)

the solution given by ronak agarwal is often used... if u don't follow his method, you can take a=1 b=-1 and c=0 or any other solution set where a+b+c=0 then also u will get "2"