Mathematics - Quadratic Equation Question with Solution | TestHub

$\therefore 3{\mathrm{x}}^6+{\mathrm{x}}^5-27{\mathrm{x}}^4+27{\mathrm{x}}^2-\mathrm{x}-3=0$\ This can be written as, $3\left({\mathrm{x}}^6-1\right)+\mathrm{x}\left({\mathrm{x}}^4-1\right)-27{\mathrm{x}}^2\left({\mathrm{x}}^2-1\right)=0$ or, $\left(x^2-1\right)\left\{3\left(x^4+x^2+1\right)+x\left(x^2+1\right)-27x^2\right\}=0$ or, $\left({\mathrm{x}}^2-1\right)\left\{3{\mathrm{x}}^4-24{\mathrm{x}}^2+{\mathrm{x}}^3+\mathrm{x}+3\right\}=0$ or, $\left({\mathrm{x}}^2-1\right)\left\{3{\mathrm{x}}^4+{\mathrm{x}}^3-24{\mathrm{x}}^2+\mathrm{x}+3\right\}=0$

So, $3{\mathrm{x}}^4+{\mathrm{x}}^3-24{\mathrm{x}}^2+\mathrm{x}+3=0$ is a reciprocal equation of even degree (i.e. 4) and first type

Hence it is standard form of reciprocal equation.