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Question Number 144839 by mathdanisur last updated on 29/Jun/21

If x = (5)^(1/3) + 3 and y = 4 (3)^(1/3) Prove that: x - y < 0

$${If}\:\:{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}}\:+\:\mathrm{3}\:\:{and}\:\:{y}\:=\:\mathrm{4}\:\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$ $${Prove}\:{that}:\:\:{x}\:-\:{y}\:<\:\mathrm{0} \\ $$

Answered by ajfour last updated on 29/Jun/21

(x−3)^3 =5 y^3 =192 ...(i) x^3 −9x^2 +27x−27=5 ...(ii) subtracting ..(i) from (ii) (x−y){[(x−y)+((3y)/2)]^2 +((3y^2 )/4)} +9x^2 +27x+160=0 for x>0, above eq. is true only if x−y<0 ; (and above eq. is of course true).

$$\left({x}−\mathrm{3}\right)^{\mathrm{3}} =\mathrm{5} \\ $$ $${y}^{\mathrm{3}} =\mathrm{192}\:\:\:...\left({i}\right) \\ $$ $${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{27}{x}−\mathrm{27}=\mathrm{5}\:\:\:...\left({ii}\right) \\ $$ $${subtracting}\:\:..\left({i}\right)\:\:{from}\:\left({ii}\right) \\ $$ $$\left({x}−{y}\right)\left\{\left[\left({x}−{y}\right)+\frac{\mathrm{3}{y}}{\mathrm{2}}\right]^{\mathrm{2}} +\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$ $$+\mathrm{9}{x}^{\mathrm{2}} +\mathrm{27}{x}+\mathrm{160}=\mathrm{0} \\ $$ $${for}\:{x}>\mathrm{0},\:\:\:{above}\:{eq}.\:{is}\:{true} \\ $$ $${only}\:{if}\:\:{x}−{y}<\mathrm{0}\:;\:\left({and}\:{above}\right. \\ $$ $$\left.{eq}.\:{is}\:{of}\:{course}\:{true}\right). \\ $$

Commented bymathdanisur last updated on 29/Jun/21

thankyou Sir cool

$${thankyou}\:{Sir}\:{cool} \\ $$

Answered by Rakshay last updated on 29/Jun/21

suppose x−y≥0 i.e ^3 (√5)+3−4^3 (√3)≥0 ⇒4^3 (√3)−^3 (√5)≤3.....(1) ⇒ (4^3 (√3)−^3 (√5))^3 ≤27 (64)(3)−(5)−3(4 )(^3 (√(15)))(4^3 (√3)−^3 (√5))≤27 204−5−12(^3 (√(15)))(4^3 (√3) −^3 (√5))≤27 ⇒172−12(^3 (√(15)))(4^3 (√3)−^3 (√5))≤0 ⇒((172)/(12))≤(^3 (√(15)))(4^3 (√3)−^3 (√5))≤(3)(^3 (√(15))) ⇒((172)/(36))≤^3 (√(15))( which isn′t true) so our supposition is wrong, thus, x−y<0

$${suppose}\:{x}−{y}\geqslant\mathrm{0} \\ $$ $${i}.{e}\:\:\:\:\:\:\:^{\mathrm{3}} \sqrt{\mathrm{5}}+\mathrm{3}−\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}\geqslant\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\Rightarrow\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}−^{\mathrm{3}} \sqrt{\mathrm{5}}\leqslant\mathrm{3}.....\left(\mathrm{1}\right) \\ $$ $$\Rightarrow\:\:\left(\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}−^{\mathrm{3}} \sqrt{\mathrm{5}}\right)^{\mathrm{3}} \leqslant\mathrm{27} \\ $$ $$\:\:\:\left(\mathrm{64}\right)\left(\mathrm{3}\right)−\left(\mathrm{5}\right)−\mathrm{3}\left(\mathrm{4}\:\right)\left(^{\mathrm{3}} \sqrt{\mathrm{15}}\right)\left(\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}−^{\mathrm{3}} \sqrt{\mathrm{5}}\right)\leqslant\mathrm{27} \\ $$ $$\mathrm{204}−\mathrm{5}−\mathrm{12}\left(^{\mathrm{3}} \sqrt{\mathrm{15}}\right)\left(\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}\:−^{\mathrm{3}} \sqrt{\mathrm{5}}\right)\leqslant\mathrm{27} \\ $$ $$\Rightarrow\mathrm{172}−\mathrm{12}\left(^{\mathrm{3}} \sqrt{\mathrm{15}}\right)\left(\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}−^{\mathrm{3}} \sqrt{\mathrm{5}}\right)\leqslant\mathrm{0} \\ $$ $$\Rightarrow\frac{\mathrm{172}}{\mathrm{12}}\leqslant\left(^{\mathrm{3}} \sqrt{\mathrm{15}}\right)\left(\mathrm{4}\:^{\mathrm{3}} \sqrt{\mathrm{3}}−^{\mathrm{3}} \sqrt{\mathrm{5}}\right)\leqslant\left(\mathrm{3}\right)\left(^{\mathrm{3}} \sqrt{\mathrm{15}}\right) \\ $$ $$\Rightarrow\frac{\mathrm{172}}{\mathrm{36}}\leqslant^{\mathrm{3}} \sqrt{\mathrm{15}}\left(\:{which}\:{isn}'{t}\:{true}\right)\: \\ $$ $${so}\:{our}\:{supposition}\:{is}\:{wrong},\:{thus}, \\ $$ $${x}−{y}<\mathrm{0} \\ $$

Commented bymathdanisur last updated on 29/Jun/21

thanks Sir cool

$${thanks}\:{Sir}\:{cool} \\ $$

Answered by mr W last updated on 29/Jun/21

(5)^(1/3) <(6)^(1/3) =((2×3))^(1/3) <((((27)/8)×3))^(1/3) =(3/2)×(3)^(1/3) 3=((9×3))^(1/3) <((((125)/8)×3))^(1/3) =(5/2)×(3)^(1/3) x=(5)^(1/3) +3<(3/2)×(3)^(1/3) +(5/2)×(3)^(1/3) =4(3)^(1/3) =y ⇒x−y<0

$$\sqrt[{\mathrm{3}}]{\mathrm{5}}<\sqrt[{\mathrm{3}}]{\mathrm{6}}=\sqrt[{\mathrm{3}}]{\mathrm{2}×\mathrm{3}}<\sqrt[{\mathrm{3}}]{\frac{\mathrm{27}}{\mathrm{8}}×\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}}×\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$ $$\mathrm{3}=\sqrt[{\mathrm{3}}]{\mathrm{9}×\mathrm{3}}<\sqrt[{\mathrm{3}}]{\frac{\mathrm{125}}{\mathrm{8}}×\mathrm{3}}=\frac{\mathrm{5}}{\mathrm{2}}×\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$ $${x}=\sqrt[{\mathrm{3}}]{\mathrm{5}}+\mathrm{3}<\frac{\mathrm{3}}{\mathrm{2}}×\sqrt[{\mathrm{3}}]{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{2}}×\sqrt[{\mathrm{3}}]{\mathrm{3}}=\mathrm{4}\sqrt[{\mathrm{3}}]{\mathrm{3}}={y} \\ $$ $$\Rightarrow{x}−{y}<\mathrm{0} \\ $$

Commented bymathdanisur last updated on 29/Jun/21

cool Sir thanks

$${cool}\:{Sir}\:{thanks} \\ $$

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