Question Number 213002 by efronzo1 last updated on 28/Oct/24
$$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/24
$$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\ $$$$\sqrt{\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}}\:=\sqrt{\mathrm{43}}\:+\mathrm{3}\:\:\: \\ $$$$\sqrt{\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}}\:=\sqrt{\mathrm{43}}\:−\mathrm{3}\:\:\: \\ $$$$\frac{\left(\sqrt{\mathrm{43}}\:+\mathrm{3}\right)^{\mathrm{3}} −\left(\sqrt{\mathrm{43}}\:−\mathrm{3}\:\right)^{\mathrm{3}} }{\mathrm{18}} \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right) \\ $$$$\frac{\left(\sqrt{\mathrm{43}}\:+\mathrm{3}−\sqrt{\mathrm{43}}\:+\mathrm{3}\:\right)\left(\:\left(\sqrt{\mathrm{43}}\:+\mathrm{3}\:\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{43}}\:+\mathrm{3}\:\right)\left(\sqrt{\mathrm{43}}\:−\mathrm{3}\right)+\left(\sqrt{\mathrm{43}}\:−\mathrm{3}\right)^{\mathrm{2}} \:\right)}{\mathrm{18}} \\ $$$$=\frac{\mathrm{6}\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:+\mathrm{43}−\mathrm{9}+\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\:\right)}{\mathrm{18}} \\ $$$$\frac{\mathrm{6}\:\left(\mathrm{52}+\mathrm{34}+\mathrm{52}\:\right)}{\mathrm{18}}=\frac{\mathrm{6}×\mathrm{138}}{\mathrm{18}}=\frac{\mathrm{138}}{\mathrm{3}}=\mathrm{46} \\ $$
Commented by golsendro last updated on 28/Oct/24
$$\:\sqrt{\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}}\:=\:\sqrt{\mathrm{43}}−\mathrm{3} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Oct/24
$$\mathcal{T}{hanks}\:{sir},\:{I}'{ve}\:{corrected}. \\ $$