An algebra problem by Gourav Kushwaha

$\frac{({x}^{4}+{x}^{2}+1)}{({x}^{2}+x+1)} = 7$

Now, we can write numerator as ${x}^{4}+2{x}^{2}+1 - {x}^{2}$

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

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

Now, the fraction can be written as -

$\frac{(({x}^{2}+1-x)({x}^{2}+1+x))}{({x}^{2}+x+1)} = 7$

${x}^{2}+1-x = 7$

${x}^{2} - x - 6 = 0$

$(x-3)(x+2) = 0$

Therefore, x = 3, -2.

Hence, SUM = 3 -2 = 1