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Question Number 159170 by mnjuly1970 last updated on 13/Nov/21

Commented by cortano last updated on 13/Nov/21
$⇒(x)^(1/3) +((a−x))^(1/3) −((2a−x))^(1/3) = 0 ⇒x+a−x−2a+x = −3((x(a−x)(2a−x)))^(1/3) ⇒x−a = −3((x(a−x)(2a−x)))^(1/3) ⇒x^3 −3ax^2 +3a^2 x−a^3 =−27(ax−x^2 )(2a−x) ⇒x^3 −3ax^2 +3a^2 x−a^3 =−27(2a^2 x−3ax^2 +x^3 ) ⇒28x^3 −84ax^2 +57a^2 x−a^3 = 0 { (α),(β),(γ) :} ⇒α×β×γ = (a^3 /(28)) = (2/7) ⇒a^3 = 8 ; a=2$
$\Rightarrow \sqrt[3]{x} + \sqrt[3]{a - x} - \sqrt[3]{2 a - x} = 0$ $\Rightarrow x + a - x - 2 a + x = - 3 \sqrt[3]{x (a - x) (2 a - x)}$ $\Rightarrow x - a = - 3 \sqrt[3]{x (a - x) (2 a - x)}$ $\Rightarrow x^{3} - 3 {a x}^{2} + 3 a^{2} x - a^{3} = - 27 (a x - x^{2}) (2 a - x)$ $\Rightarrow x^{3} - 3 {a x}^{2} + 3 a^{2} x - a^{3} = - 27 (2 a^{2} x - 3 {a x}^{2} + x^{3})$ $\Rightarrow 28 x^{3} - 84 {a x}^{2} + 57 a^{2} x - a^{3} = 0 {\begin{cases} α \\ β \\ γ \end{cases}$ $\Rightarrow α \times β \times γ = \frac{a^{3}}{28} = \frac{2}{7}$ $\Rightarrow a^{3} = 8; a = 2$
	Answered by Rasheed.Sindhi last updated on 14/Nov/21
	$(x)^(1/3) +((a−x))^(1/3) =((2a−x))^(1/3) ((x)^(1/3) +((a−x))^(1/3) )^3 =(((2a−x))^(1/3) )^3 x+(a−x)+3((x)^(1/3) ((a−x))^(1/3) )((x)^(1/3) +((a−x))^(1/3) ) =2a−x 3((x)^(1/3) ((a−x))^(1/3) )((2a−x))^(1/3) =a−x 27x(a−x)(2a−x)=(a−x)^3 (a−x)^3 −27x(a−x)(2a−x)=0 (a−x){(a−x)^2 −27x(2a−x)}=0 (a−x){28x^2 −56ax+a^2 }=0 x=a_(a=α (say)) ∣ 28x^2 −56ax+a^2 }=0_(βγ=a^2 /28) αβγ=a((a^2 /(28)))=(2/7)⇒a^3 =8⇒a=2$
	$\sqrt[3]{x} + \sqrt[3]{a - x} = \sqrt[3]{2 a - x}$ ${(\sqrt[3]{x} + \sqrt[3]{a - x})}^{3} = {(\sqrt[3]{2 a - x})}^{3}$ $x + (a - x) + 3 (\sqrt[3]{x} \sqrt[3]{a - x}) (\sqrt[3]{x} + \sqrt[3]{a - x})$ $= 2 a - x$ $3 (\sqrt[3]{x} \sqrt[3]{a - x}) \sqrt[3]{2 a - x} = a - x$ $27 x (a - x) (2 a - x) = {(a - x)}^{3}$ ${(a - x)}^{3} - 27 x (a - x) (2 a - x) = 0$ $(a - x) {{(a - x)}^{2} - 27 x (2 a - x)} = 0$ $(a - x) {28 x^{2} - 56 a x + a^{2}} = 0$ $Missing \left or extra \right$ $α β γ = a (\frac{a^{2}}{28}) = \frac{2}{7} \Rightarrow a^{3} = 8 \Rightarrow a = 2$
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