If-x-5-1-3-3-and-y-4-3-1-3-Prove-that-x-y-lt-0-

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 \dots (i)

x^{3} - 9 x^{2} + 27 x - 27 = 5 \dots (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 by mathdanisur 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 \dots . . (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 by mathdanisur 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 by mathdanisur last updated on 29/Jun/21

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