how to solve ((x−1))^(1/(3 )) + ((x−3))^(1/(3 )) + ((x−5))^(1/(3 )) = 0


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Question Number 85822 by jagoll last updated on 25/Mar/20

$how to solve$ $\sqrt[3]{x - 1} + \sqrt[3]{x - 3} + \sqrt[3]{x - 5} = 0$
Commented by jagoll last updated on 25/Mar/20

$thank mr$
	Answered by john santu last updated on 25/Mar/20

	Answered by TANMAY PANACEA. last updated on 25/Mar/20
	$when a+b+c=0 a^3 +b^3 +c^3 =3abc (x−1)+(x−3)+(x−5)=3(x−1)^(1/3) (x−3)^(1/3) (x−5)^(1/3) 3(x−3)=3{(x−1)(x−3)(x−5)}^(1/3) (x−3)^3 =(x−1)(x−3)(x−5) (x−3){x^2 −6x+9−(x−1)(x−5)}=0 (x−3)(x^2 −6x+9−x^2 +6x−5)=0 x=3$
	$w h e n a + b + c = 0$ $a^{3} + b^{3} + c^{3} = 3 a b c$ $(x - 1) + (x - 3) + (x - 5) = 3 {(x - 1)}^{\frac{1}{3}} {(x - 3)}^{\frac{1}{3}} {(x - 5)}^{\frac{1}{3}}$ $3 (x - 3) = 3 {(x - 1) (x - 3) (x - 5)}^{\frac{1}{3}}$ ${(x - 3)}^{3} = (x - 1) (x - 3) (x - 5)$ $(x - 3) {x^{2} - 6 x + 9 - (x - 1) (x - 5)} = 0$ $(x - 3) (x^{2} - 6 x + 9 - x^{2} + 6 x - 5) = 0$ $x = 3$
	Commented by jagoll last updated on 25/Mar/20

	$thank mr$
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