Question Number 10572 by j.masanja06@gmail.com last updated on 18/Feb/17
$$\mathrm{factorise} \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1} \\ $$
Answered by sandy_suhendra last updated on 18/Feb/17
$$=\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right) \\ $$
Commented by FilupS last updated on 19/Feb/17
$${x}^{\mathrm{3}} +\mathrm{1}={x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −{x}+{x}+\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{1}={x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)+{x}\left({x}−\mathrm{1}\right)+{x}+\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{1}=??? \\ $$$${x}^{\mathrm{3}} +\mathrm{1}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\mathrm{Please}\:\mathrm{help} \\ $$
Commented by prakash jain last updated on 19/Feb/17
$${x}^{\mathrm{3}} +\mathrm{1}={x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}^{\mathrm{2}} −{x}+{x}+\mathrm{1} \\ $$$$={x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)−{x}\left({x}+\mathrm{1}\right)+\mathrm{1}\left({x}+\mathrm{1}\right) \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$
Commented by FilupS last updated on 19/Feb/17
$$\mathrm{ahh}\:\mathrm{I}\:\mathrm{grouped}\:\mathrm{the}\:\mathrm{wrong}\:\mathrm{terms} \\ $$$$\: \\ $$$$\mathrm{thanks} \\ $$