Fraction solving related

Question

If $$ a + b + c = 0 $$ , then evaluate $$\frac{a^4 + b^4 + c^4}{a^2b^2 + c^2(a^2 + b^2)}$$

I tried solving but could make to mentionworthy progress.

Answer 1

$a+b=-c , a^2+b^2=c^2-2ab$

$$\frac{a^4+b^4+c^4} {a^2b^2+c^2(a^2+b^2)} \iff \frac{(a^2+b^2)^2-2a^2b^2+c^4}{a^2b^2+c^2(a^2+b^2)}\iff \frac {(c^2-2ab)^2-2a^2b^2+c^4}{a^2b^2+c^2(c^2-2ab)}\iff \frac {c^4-4abc^2+4a^2b^2-2a^2b^2+c^4}{c^4-2abc^2+a^2b^2}\iff \frac{2c^4-4abc^2+2a^2b^2}{c^4-2abc^2+a^2b^2}=2$$

Answer 2

this is $$\frac{a^4+b^4+(-a-b)^4}{a^2b^2+(-a-b)^2(a^2+b^2)}=2$$ after simplifying

Answer 3

Since $$\sum_{cyc}(2a^2b^2-a^4)=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$ $$= (2ab)^2-(a^2+b^2-c^2)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0,$$ we obtain $$\frac{a^4+b^4+c^4}{a^2b^2+a^2c^2+b^2c^2}=2.$$

Answer 4

If $$a+b+c=0\;\;\;/^2$$ then $$a^2+b^2+c^2 = -2(ab+bc+ca)\;\;\;/^2$$ and then $$a^4+b^4+c^4 +2(a^2b^2+b^2c^2+c^2a^2)= 4(a^2b^2+a^2b^2c+c^2a^2)+8abc\underbrace{(a+b+c)}_{=0}$$ so $$a^4+b^4+c^4 = 2(a^2b^2+a^2b^2c+c^2a^2)$$ and so ...

Answer 5

Well, we have the identity $$ \left( a+b+c \right) ^{4}-6\,({a}^{2}{b}^{2}+\,{a}^{2}{c}^{2}+\,{b}^{2}{c}^{2})-{a}^{4}-{b}^{4}-{c}^{4}=4(a^3(b+c)+b^3(a+c)+c^3(a+b))+12 abc (a+b+c). $$ Since $a+b+c=0$ we obtain $$ -6\,({a}^{2}{b}^{2}+\,{a}^{2}{c}^{2}+\,{b}^{2}{c}^{2})-{a}^{4}-{b}^{4}-{c}^{4}=4(a^3(b+c)+b^3(a+c)+c^3(a+b))=-4 (a^4+b^4+c^4). $$ Thus $$ 6\,({a}^{2}{b}^{2}+\,{a}^{2}{c}^{2}+\,{b}^{2}{c}^{2})=3 (a^4+b^4+c^4), $$ and $$ \frac{a^4+b^4+c^4}{{a}^{2}{b}^{2}+\,{a}^{2}{c}^{2}+\,{b}^{2}{c}^{2}}=2. $$

Answer 6

Consider $a+b+c=S$, $ab+ac+bc=Q$, $abc=P$. By the theory of symmetric polynomials, we can write $$ a^4+b^4+c^4=tS^4+xS^2Q+yQ^2+zSP $$ and we can determine the coefficients by using special values: \begin{align} &a=1,b=0,c=0 & 1&=t \\ &a=1,b=1,c=0 & 2&=16t+4x+y \\ &a=1,b=1,c=1 & 3&=81t+27x+9y+3z \\ &a=1,b=-1,c=0 & 2&=y \end{align} so $t=1$, $x=-4$, $y=2$, $z=4$.

Similarly $a^2b^2+a^2c^2+b^2c^2=tS^4+xS^2Q+yQ^2+zSP$ and \begin{align} &a=1,b=0,c=0 & 0&=t \\ &a=1,b=1,c=0 & 1&=16t+4x+y \\ &a=1,b=1,c=1 & 3&=81t+27x+9y+3z \\ &a=2,b=1,c=0 & 4&=81t+18x+4y \end{align} so $t=0$, $x=0$, $y=1$ and $z=-2$. Hence, since in your case $S=0$, we have $$ \frac{a^4 + b^4 + c^4}{a^2b^2 + c^2(a^2 + b^2)} = \frac{S^4-4S^2Q+2Q^2+4SP}{Q^2-2SP}=\frac{2Q^2}{Q^2}=2 $$