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Question Number 204658 by hardmath last updated on 24/Feb/24
If a = (9)^(1/3) − (3)^(1/3) + 1 Find (((4 − a)/a))^6 = ?
$$\mathrm{If}\:\:\:\mathrm{a}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\mathrm{1} \\ $$$$\mathrm{Find}\:\:\:\left(\frac{\mathrm{4}\:−\:\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{6}} =\:? \\ $$
Answered by Rasheed.Sindhi last updated on 24/Feb/24
a = (9)^(1/3) − (3)^(1/3) + 1; (((4 − a)/a))^6 = ? let (3)^(1/3) =t⇒t^3 =3 a=t^2 −t+1 (((4 − a)/a))^6 =((4/a)−1)^6 =(((3+1)/a)−1)^6 =(((t^3 +1)/( t^2 −t+1))−1)^6 =((((t+1)( t^2 −t+1))/( t^2 −t+1))−1)^6 =(t+1−1)^6 =(t^3 )^2 =3^2 =9
$$\:\mathrm{a}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\mathrm{1};\:\left(\frac{\mathrm{4}\:−\:\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{6}} =\:? \\ $$$$\mathrm{let}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:=\mathrm{t}\Rightarrow\mathrm{t}^{\mathrm{3}} =\mathrm{3} \\ $$$$\mathrm{a}=\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1} \\ $$$$\left(\frac{\mathrm{4}\:−\:\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{6}} =\left(\frac{\mathrm{4}}{\mathrm{a}}−\mathrm{1}\right)^{\mathrm{6}} =\left(\frac{\mathrm{3}+\mathrm{1}}{\mathrm{a}}−\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{t}^{\mathrm{3}} +\mathrm{1}}{\:\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}−\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\left(\mathrm{t}+\mathrm{1}\right)\cancel{\left(\:\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}\right)}}{\cancel{\:\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}}−\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{t}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{t}^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$
Answered by A5T last updated on 24/Feb/24
Let p=(3)^(1/3) ; a(p+1)=p^3 +1=4⇒4−a=a(p+1)−a ⇒4−a=ap⇒(((4−a)/a))^6 =p^6 =3^2 =9
$${Let}\:{p}=\sqrt[{\mathrm{3}}]{\mathrm{3}};\:{a}\left({p}+\mathrm{1}\right)={p}^{\mathrm{3}} +\mathrm{1}=\mathrm{4}\Rightarrow\mathrm{4}−{a}={a}\left({p}+\mathrm{1}\right)−{a} \\ $$$$\Rightarrow\mathrm{4}−{a}={ap}\Rightarrow\left(\frac{\mathrm{4}−{a}}{{a}}\right)^{\mathrm{6}} ={p}^{\mathrm{6}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$

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