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Question Number 161505 by naka3546 last updated on 18/Dec/21
((a + ((a+8)/3) (√((a−1)/3))))^(1/3) + ((a − ((a+8)/3) (√((a−1)/3))))^(1/3) = ?
$$\sqrt[{\mathrm{3}}]{{a}\:+\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:+\:\sqrt[{\mathrm{3}}]{{a}\:−\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:\:=\:\:? \\ $$
Commented by cortano last updated on 18/Dec/21
x=((a+(((a+8))/3)(√((a−1)/3))))^(1/3) +((a−(((a+8))/3)(√((a−1)/3))))^(1/3) ⇒(a+(((a+8)/3))(√((a−1)/3)))+(a−(((a+8)/3))(√((a−1)/3)))−x^3 =−3x((a^2 −(((a+8)/3))^2 (((a−1)/3))))^(1/3) ⇒2a−x^3 =−3x (((27a^2 −(a−1)(a+8)^2 )/(27)))^(1/3) ⇒2a−x^3 =−x ((27a^2 −(a^2 +16a+64)(a−1)))^(1/3) ⇒2a−x^3 =−x ((27a^2 −(a^3 +15a^2 +48a−64)))^(1/3) ⇒2a−x^3 =−x ((−a^3 +12a^2 −48a+64))^(1/3) ⇒2a−x^3 −x (((4−a)^3 ))^(1/3) ⇒2a−x^3 = −x(4−a) ⇒x^3 −(4−a)x−2a=0 Use Cardano
$$\:{x}=\sqrt[{\mathrm{3}}]{{a}+\frac{\left({a}+\mathrm{8}\right)}{\mathrm{3}}\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:+\sqrt[{\mathrm{3}}]{{a}−\frac{\left({a}+\mathrm{8}\right)}{\mathrm{3}}\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}} \\ $$$$\:\Rightarrow\left({a}+\cancel{\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\right)+\left({a}−\cancel{\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\right)−{x}^{\mathrm{3}} =−\mathrm{3}{x}\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} −\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\frac{{a}−\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} \:=−\mathrm{3}{x}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{27}{a}^{\mathrm{2}} −\left({a}−\mathrm{1}\right)\left({a}+\mathrm{8}\right)^{\mathrm{2}} }{\mathrm{27}}} \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} \:=−{x}\:\sqrt[{\mathrm{3}}]{\mathrm{27}{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +\mathrm{16}{a}+\mathrm{64}\right)\left({a}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} \:=−{x}\:\sqrt[{\mathrm{3}}]{\mathrm{27}{a}^{\mathrm{2}} −\left({a}^{\mathrm{3}} +\mathrm{15}{a}^{\mathrm{2}} +\mathrm{48}{a}−\mathrm{64}\right)} \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} \:=−{x}\:\sqrt[{\mathrm{3}}]{−{a}^{\mathrm{3}} +\mathrm{12}{a}^{\mathrm{2}} −\mathrm{48}{a}+\mathrm{64}} \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} \:−{x}\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−{a}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\mathrm{2}{a}−{x}^{\mathrm{3}} =\:−{x}\left(\mathrm{4}−{a}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\left(\mathrm{4}−{a}\right){x}−\mathrm{2}{a}=\mathrm{0} \\ $$$$\:{Use}\:{Cardano}\: \\ $$
Commented by Ar Brandon last updated on 19/Dec/21
That takes you back to the start.
$$\mathrm{That}\:\mathrm{takes}\:\mathrm{you}\:\mathrm{back}\:\mathrm{to}\:\mathrm{the}\:\mathrm{start}. \\ $$
Commented by naka3546 last updated on 19/Dec/21
Thank you, sir.
$${Thank}\:\:{you},\:\:{sir}. \\ $$
Answered by MJS_new last updated on 19/Dec/21
p^(1/3) +q^(1/3) =x (p^(1/3) +q^(1/3) )^3 =x^3 [might introduce false solutions] p+3p^(1/3) q^(1/3) (p^(1/3) +q^(1/3) )+q=x^3 p+3p^(1/3) q^(1/3) x+q=x^3 3p^(1/3) q^(1/3) x=x^3 −p−q 27pqx^3 =(x^3 −p−q)^3 p=u+v∧q=u−v 27(u^2 −v^2 )x^3 =(x^3 −2u)^3 u=a∧v=((a+8)/3)(√((a−1)/3)) (4−a)^3 x^3 =(x^3 −2a)^3 (4−a)x=x^3 −2a x^3 +(a−4)x−2a=0 (x−2)(x^2 +2x+a)=0 I think this is the only valid solution: x=2
$${p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} ={x} \\ $$$$\left({p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} \right)^{\mathrm{3}} ={x}^{\mathrm{3}} \:\:\:\:\:\left[\mathrm{might}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}\right] \\ $$$${p}+\mathrm{3}{p}^{\mathrm{1}/\mathrm{3}} {q}^{\mathrm{1}/\mathrm{3}} \left({p}^{\mathrm{1}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} \right)+{q}={x}^{\mathrm{3}} \\ $$$${p}+\mathrm{3}{p}^{\mathrm{1}/\mathrm{3}} {q}^{\mathrm{1}/\mathrm{3}} {x}+{q}={x}^{\mathrm{3}} \\ $$$$\mathrm{3}{p}^{\mathrm{1}/\mathrm{3}} {q}^{\mathrm{1}/\mathrm{3}} {x}={x}^{\mathrm{3}} −{p}−{q} \\ $$$$\mathrm{27}{pqx}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −{p}−{q}\right)^{\mathrm{3}} \\ $$$${p}={u}+{v}\wedge{q}={u}−{v} \\ $$$$\mathrm{27}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right){x}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −\mathrm{2}{u}\right)^{\mathrm{3}} \\ $$$${u}={a}\wedge{v}=\frac{{a}+\mathrm{8}}{\mathrm{3}}\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{4}−{a}\right)^{\mathrm{3}} {x}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −\mathrm{2}{a}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{4}−{a}\right){x}={x}^{\mathrm{3}} −\mathrm{2}{a} \\ $$$${x}^{\mathrm{3}} +\left({a}−\mathrm{4}\right){x}−\mathrm{2}{a}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+{a}\right)=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{valid}\:\mathrm{solution}:\:{x}=\mathrm{2} \\ $$
Commented by naka3546 last updated on 19/Dec/21
Thank you, sir.
$${Thank}\:\:{you},\:\:{sir}. \\ $$
Commented by Tawa11 last updated on 19/Dec/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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