If x = ((2019))^(1/3) + 1 Find: (x + 1)^3 −6∙(x + 1)^2 + 12x − 3


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Question Number 203245 by hardmath last updated on 13/Jan/24

$If x = \sqrt[3]{2019} + 1$ $Find :$ ${(x + 1)}^{3} - 6 \cdot {(x + 1)}^{2} + 12 x - 3$
	Answered by mr W last updated on 13/Jan/24

	$s a y t = x - 1 = \sqrt[3]{2019}$ $\Rightarrow t^{3} = 2019$ ${(x + 1)}^{3} - 6 {(x + 1)}^{2} + 12 x + 9$ $= {(x - 1 + 2)}^{3} - 6 {(x - 1 + 2)}^{2} + 12 (x - 1 + 1) + 9$ $= {(t + 2)}^{3} - 6 {(t + 2)}^{2} + 12 (t + 1) + 9$ $= t^{3} + 6 t^{2} + 12 t + 8 - 6 (t^{2} + 4 t + 4) + 12 t + 12 + 9$ $= t^{3} + 5$ $= 2019 + 5$ $= 2024$
	Commented by hardmath last updated on 13/Jan/24

	$Dear professor, answer : 2012$
	Commented by mr W last updated on 13/Jan/24

	${c a n}^{'} t y o u s e e t h a t i h a v e c h a n g e d t h e$ $q u e s t i o n p u r p o s e l y t o + 9 i n o r d e r t o$ $g e t a n i c e 2024 i n s t e a d o f 2012 ?$
	Answered by Rasheed.Sindhi last updated on 13/Jan/24

	${(x + 1)}^{3} - 6 {(x + 1)}^{2} + 12 (x + 1) - 15$ $l e t x + 1 = y$ $y^{3} - 6 y^{2} + 12 y - 15$ $y^{3} - 6 y (y - 2) - 15$ $y^{3} - 3 y (y - 2) - 8 - 7$ ${(y - 2)}^{3} - 7$ ${(x + 1 - 2)}^{3} - 7$ ${(\sqrt[3]{2019} + 1 - 1)}^{3} - 7$ $2019 - 7 = 2012$
	Commented by Rasheed.Sindhi last updated on 13/Jan/24

	$M a y b e t h e b o o k, i n w h i c h t h e$ $a n s w e r 2012 i s w r i t t e n, w a s$ $p u b l i s h e d i n 2012 a n d b y t h e n$ $2012 w a s a s n i c e a s 2024$ $t o d a y . :)$
	Commented by hardmath last updated on 13/Jan/24

	$My dear professors,$ I think years are not important)
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