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


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

$I f x = \sqrt[3]{5} + 3 a n d y = 4 \sqrt[3]{3}$ $P r o v e t h a t : x - y < 0$
	Answered by ajfour last updated on 29/Jun/21
	$(x−3)^3 =5 y^3 =192 ...(i) x^3 −9x^2 +27x−27=5 ...(ii) subtracting ..(i) from (ii) (x−y){[(x−y)+((3y)/2)]^2 +((3y^2 )/4)} +9x^2 +27x+160=0 for x>0, above eq. is true only if x−y<0 ; (and above eq. is of course true).$
	${(x - 3)}^{3} = 5$ $y^{3} = 192 . . . (i)$ $x^{3} - 9 x^{2} + 27 x - 27 = 5 . . . (i i)$ $s u b t r a c t i n g . . (i) f r o m (i i)$ $(x - y) {{[(x - y) + \frac{3 y}{2}]}^{2} + \frac{3 y^{2}}{4}}$ $+ 9 x^{2} + 27 x + 160 = 0$ $f o r x > 0, a b o v e e q . i s t r u e$ $o n l y i f x - y < 0; (a n d a b o v e$ $e q . i s o f c o u r s e t r u e) .$
	Commented bymathdanisur last updated on 29/Jun/21

	$t h a n k y o u S i r c o o l$
	Answered by Rakshay last updated on 29/Jun/21

	$s u p p o s e x - y ⩾ 0$ $i . e^{3} \sqrt{5} + 3 - 4^{3} \sqrt{3} ⩾ 0$ $\Rightarrow 4^{3} \sqrt{3} -^{3} \sqrt{5} ⩽ 3 . . . . . (1)$ $\Rightarrow {(4^{3} \sqrt{3} -^{3} \sqrt{5})}^{3} ⩽ 27$ $(64) (3) - (5) - 3 (4) (^{3} \sqrt{15}) (4^{3} \sqrt{3} -^{3} \sqrt{5}) ⩽ 27$ $204 - 5 - 12 (^{3} \sqrt{15}) (4^{3} \sqrt{3} -^{3} \sqrt{5}) ⩽ 27$ $\Rightarrow 172 - 12 (^{3} \sqrt{15}) (4^{3} \sqrt{3} -^{3} \sqrt{5}) ⩽ 0$ $\Rightarrow \frac{172}{12} ⩽ (^{3} \sqrt{15}) (4^{3} \sqrt{3} -^{3} \sqrt{5}) ⩽ (3) (^{3} \sqrt{15})$ $\Rightarrow \frac{172}{36} ⩽^{3} \sqrt{15} (w h i c h {i s n}^{'} t t r u e)$ $s o o u r s u p p o s i t i o n i s w r o n g, t h u s,$ $x - y < 0$
	Commented bymathdanisur last updated on 29/Jun/21

	$t h a n k s S i r c o o l$
	Answered by mr W last updated on 29/Jun/21

	$\sqrt[3]{5} < \sqrt[3]{6} = \sqrt[3]{2 \times 3} < \sqrt[3]{\frac{27}{8} \times 3} = \frac{3}{2} \times \sqrt[3]{3}$ $3 = \sqrt[3]{9 \times 3} < \sqrt[3]{\frac{125}{8} \times 3} = \frac{5}{2} \times \sqrt[3]{3}$ $x = \sqrt[3]{5} + 3 < \frac{3}{2} \times \sqrt[3]{3} + \frac{5}{2} \times \sqrt[3]{3} = 4 \sqrt[3]{3} = y$ $\Rightarrow x - y < 0$
	Commented bymathdanisur last updated on 29/Jun/21

	$c o o l S i r t h a n k s$
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