$\frac{79 ^4+79 ^2+1}{79 ^2+79+1}=80 ^2-A$

Let's assume: $x=79$

Then, taking LHS we get:

$\frac{79^{4}+79^{2}+1}{79^{2}+79+1} = \frac{x^{4}+x^{2}+1}{x^{2}+x+1}$

So, factorizing the numerator we get:

$x^{4}+x^{2}+1=(x^{4}+2x^{2}+1)-x^{2}$

$=(x^{2}+1)^{2}-(x)^{2}=(x^{2}+x+1)(x^{2}-x+1)$

So, $x^{4}+x^{2}+1=(x^{2}+x+1)(x^{2}-x+1)$

So, $\frac{x^{4}+x^{2}+1}{x^{2}+x+1}=\frac{(x^{2}+x+1)(x^{2}-x+1)}{x^{2}+x+1}$

Canceling $(x^{2}+x+1)$ , we get:

$\frac{(x^{2}+x+1)(x^{2}-x+1)}{x^{2}+x+1}=x^{2}-x+1$

$x^{2}-x+1=79^{2}-79+1=79^{2}-78$

Now, the equation is: $79^{2}-78=80^{2}-A$

$A-78=80^{2}-79^{2}$

$A-78=(80+79)(80-79)$

$A-78=159\times1$

$A=159+78$

$A=237$

So, the answer is: $A=\boxed{237}$

Considering only the LHS and taking units digit only we get [(1+1+1)/(1+9+1)] = 3/1 = 3. Hence the LHS is 3. RHS must be equal to LHS. Comsidering RHS gives us units digit of 80^2 = 0. Hence to get RHS = 3; The only number possible is one having units digit 7. (Since 80^2 - 7 = Units digit 3) Hence the answer is 237.

formula is (x^4+X^2+1)=(x^2+x+1)(x^2-x+1) where x=79 put on formula get (79^2-79+1)=80^2-A after simplification can get a=237