Question Number 36119 by anuyadavdeepak@gmail.com last updated on 29/May/18
$$\mathrm{3}\sqrt{\mathrm{43}\hat {\mathrm{3}}} \\ $$
Commented by Joel579 last updated on 29/May/18
$$\mathrm{since}\:\mathrm{433}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{number},\:\mathrm{it}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{simplified} \\ $$
Commented by anuyadavdeepak@gmail.com last updated on 29/May/18
$${The}\:\mathrm{3}\:{is}\:\:{the}\:{power}\:{of}\:\mathrm{43}….\:{sorry}\:…{i}\:{have}\:{marked}\:{with}\:{a}\:{upper}\:{sign}\:\:{if}\:{u}\:{can}\:<{see}> \\ $$
Commented by Joel579 last updated on 29/May/18
$$\mathrm{3}\sqrt{\left(\mathrm{43}^{\mathrm{3}} \right)}\:\mathrm{did}\:\mathrm{u}\:\mathrm{mean}\:\mathrm{this}? \\ $$$$ \\ $$$$\mathrm{3}\sqrt{\mathrm{43}^{\mathrm{2}} \:.\:\mathrm{43}}\:=\:\left(\mathrm{3}\:.\:\mathrm{43}\right)\sqrt{\mathrm{43}}\:=\:\mathrm{129}\sqrt{\mathrm{43}} \\ $$