Question Number 25899 by maheshagarwal last updated on 16/Dec/17
$$\mathrm{8}{a}^{\mathrm{6}} +\mathrm{5}{a}^{\mathrm{3}} +\mathrm{1}={factorize} \\ $$
Commented by sushmitak last updated on 16/Dec/17
$${a}^{\mathrm{3}} ={x} \\ $$$${x}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{32}}}{\mathrm{16}}=\frac{−\mathrm{5}\pm\iota\sqrt{\mathrm{7}}}{\mathrm{16}} \\ $$$${factors} \\ $$$$\left({a}^{\mathrm{3}} −\frac{−\mathrm{5}+\iota\sqrt{\mathrm{7}}}{\mathrm{16}}\right)\left({a}^{\mathrm{3}} −\frac{−\mathrm{5}−\iota\sqrt{\mathrm{7}}}{\mathrm{16}}\right) \\ $$
Answered by maheshagarwal last updated on 16/Dec/17
