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Question Number 147673 by mnjuly1970 last updated on 22/Jul/21
Answered by Rasheed.Sindhi last updated on 22/Jul/21
(x)^(1/3) +((x−2))^(1/3) =((x−1))^(1/3) ((x)^(1/3) +((x−2))^(1/3) )^3 =(((x−1))^(1/3) )^3 x+x−2+3(x)^(1/3) ((x−2))^(1/3) ((x)^(1/3) +((x−2))^(1/3) )=x−1 3(x)^(1/3) ((x−2))^(1/3) (((x−1))^(1/3) )=−x+1 27(x)(x−1)(x−2)=−(x−1)^3 27(x)(x−1)(x−2)+(x−1)^3 =0 (x−1){27(x)(x−2)+(x−1)^2 }=0 x=1 ∣ 27(x)(x−2)+(x−1)^2 =0 27x^2 −54x+x^2 −2x+1=0 28x^2 −56x+1=0 x=((56±(√(56^2 −4(28))))/(56)) =((56±12(√(21)))/(56))=((14±3(√(21)))/(14)) a=1,b=((14+3(√(21)))/(14)),c=((14−3(√(21)))/(14)) ((a/(a−2)))^(1/3) =((1/(1−2)))^(1/3) =−1 ((b/(b−2)))^(1/3) =((((14+3(√(21)))/(14))/(((14+3(√(21)))/(14))−2)))^(1/3) =(((14+3(√(21)))/(14+3(√(21))−28)))^(1/3) =(((14+3(√(21)))/(−14+3(√(21)))))^(1/3) =(((14+3(√(21)))^(2/3) )/({196−9(21)}^(1/3) )) =(((14+3(√(21)))^(2/3) )/({196−9(21)}^(1/3) ))=(((14+3(√(21)))^(2/3) )/7^(1/3) ) ((c/(c−2)))^(1/3) =(((14−3(√(21)))/(−14−3(√(21)))))^(1/3) =(((14−3(√(21)))^(2/3) )/({196−9(21)}^(1/3) ))=(((14−3(√(21)))^(2/3) )/7^(1/3) ) ((a/(a−2)))^(1/3) +((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) =−1+(((14+3(√(21)))^(2/3) )/7^(1/3) )+(((14−3(√(21)))^(2/3) )/7^(1/3) ) =−1+(((385+84(√(21)))^(1/3) )/7^(1/3) )+(((385−84(√(21)))^(1/3) )/7^(1/3) )
$$\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}=\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\right)^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\right)^{\mathrm{3}} \\ $$$${x}+{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\right)={x}−\mathrm{1} \\ $$$$\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\right)=−{x}+\mathrm{1} \\ $$$$\mathrm{27}\left({x}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)=−\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\mathrm{27}\left({x}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left\{\mathrm{27}\left({x}\right)\left({x}−\mathrm{2}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\mid\:\mathrm{27}\left({x}\right)\left({x}−\mathrm{2}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{27}{x}^{\mathrm{2}} −\mathrm{54}{x}+{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{28}{x}^{\mathrm{2}} −\mathrm{56}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{56}\pm\sqrt{\mathrm{56}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{28}\right)}}{\mathrm{56}} \\ $$$$\:\:\:\:=\frac{\mathrm{56}\pm\mathrm{12}\sqrt{\mathrm{21}}}{\mathrm{56}}=\frac{\mathrm{14}\pm\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}} \\ $$$${a}=\mathrm{1},{b}=\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}},{c}=\frac{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{a}}{{a}−\mathrm{2}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}=\sqrt[{\mathrm{3}}]{\frac{\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}}{\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}−\mathrm{2}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}−\mathrm{28}}} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{−\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}}=\frac{\left(\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\left\{\mathrm{196}−\mathrm{9}\left(\mathrm{21}\right)\right\}^{\mathrm{1}/\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\left\{\mathrm{196}−\mathrm{9}\left(\mathrm{21}\right)\right\}^{\mathrm{1}/\mathrm{3}} }=\frac{\left(\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}{−\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}} \\ $$$$=\frac{\left(\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\left\{\mathrm{196}−\mathrm{9}\left(\mathrm{21}\right)\right\}^{\mathrm{1}/\mathrm{3}} }=\frac{\left(\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{a}}{{a}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}} \\ $$$$=−\mathrm{1}+\frac{\left(\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} }+\frac{\left(\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$$$=−\mathrm{1}+\frac{\left(\mathrm{385}+\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} }+\frac{\left(\mathrm{385}−\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$
Commented by mr W last updated on 22/Jul/21
=(((14+3(√(21)))/(−14+3(√(21)))))^(1/3) =(((14+3(√(21)))^(2/3) )/({−196+9(21)}^(1/3) )) .... =−1−(((385+84(√(21)))^(1/3) )/7^(1/3) )−(((385−84(√(21)))^(1/3) )/7^(1/3) ) =−6
$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{−\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}}=\frac{\left(\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}\right)^{\mathrm{2}/\mathrm{3}} }{\left\{−\mathrm{196}+\mathrm{9}\left(\mathrm{21}\right)\right\}^{\mathrm{1}/\mathrm{3}} } \\ $$$$…. \\ $$$$=−\mathrm{1}−\frac{\left(\mathrm{385}+\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} }−\frac{\left(\mathrm{385}−\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$$$=−\mathrm{6} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Jul/21
Sir how′s the middle line equal to −6?
$$\boldsymbol{\mathrm{Sir}}\:\mathrm{how}'\mathrm{s}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{line}\:\mathrm{equal}\:\mathrm{to}\:−\mathrm{6}? \\ $$
Commented by mr W last updated on 22/Jul/21
=−1−(((385+84(√(21)))^(1/3) )/7^(1/3) )−(((385−84(√(21)))^(1/3) )/7^(1/3) ) =−1−(55+12(√(21)))^(1/3) −(55−12(√(21)))^(1/3) a=(55+12(√(21)))^(1/3) b=(55−12(√(21)))^(1/3) a^3 +b^3 =110 ab=1 (a+b)^3 =a^3 +b^3 +3ab(a+b) (a+b)^3 =110+3(a+b) (a+b)^3 −3(a+b)−110=0 u=a+b (u−5)(u^2 +5u+22)=0 ⇒a+b=u=5 ⇒(55+12(√(21)))^(1/3) +(55−12(√(21)))^(1/3) =5
$$=−\mathrm{1}−\frac{\left(\mathrm{385}+\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} }−\frac{\left(\mathrm{385}−\mathrm{84}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{7}^{\mathrm{1}/\mathrm{3}} } \\ $$$$=−\mathrm{1}−\left(\mathrm{55}+\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{55}−\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${a}=\left(\mathrm{55}+\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${b}=\left(\mathrm{55}−\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{110} \\ $$$${ab}=\mathrm{1} \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} =\mathrm{110}+\mathrm{3}\left({a}+{b}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}\left({a}+{b}\right)−\mathrm{110}=\mathrm{0} \\ $$$${u}={a}+{b} \\ $$$$\left({u}−\mathrm{5}\right)\left({u}^{\mathrm{2}} +\mathrm{5}{u}+\mathrm{22}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}={u}=\mathrm{5} \\ $$$$\Rightarrow\left(\mathrm{55}+\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{55}−\mathrm{12}\sqrt{\mathrm{21}}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{5} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jul/21
A bundle of thanks sir!
$$\mathrm{A}\:\mathrm{bundle}\:\mathrm{of}\:\mathrm{thanks}\:\mathrm{sir}! \\ $$
Commented by mnjuly1970 last updated on 23/Jul/21
grateful...
$${grateful}… \\ $$
Answered by mr W last updated on 22/Jul/21
((x/(x−2)))^(1/3) +1=(((x−1)/(x−2)))^(1/3) ((x/(x−2)))^(1/3) +1=(((x−1)/x))^(1/3) ((x/(x−2)))^(1/3) ((((x−1)/x))^(1/3) −1)((x/(x−2)))^(1/3) =1 let t=((x/(x−2)))^(1/3) ⇒x=((2t^3 )/(t^3 −1)) ⇒x−1=((t^3 +1)/(t^3 −1)) ((((t^3 +1)/(2t^3 )))^(1/3) −1)t=1 (((t^3 +1)/2))^(1/3) =t+1 ⇒t^3 +1=2(t^3 +3t^2 +3t+1) ⇒t^3 +6t^2 +6t+1=0 ⇒Σ_(i=1) ^3 t_i =−6 i.e. Σ_(i=1) ^3 ((x_i /(x_i −2)))^(1/3) =−6 i.e. ((a/(a−2)))^(1/3) +((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) =−6
$$\sqrt[{\mathrm{3}}]{\frac{{x}}{{x}−\mathrm{2}}}+\mathrm{1}=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{x}}{{x}−\mathrm{2}}}+\mathrm{1}=\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}}}\sqrt[{\mathrm{3}}]{\frac{{x}}{{x}−\mathrm{2}}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}}}−\mathrm{1}\right)\sqrt[{\mathrm{3}}]{\frac{{x}}{{x}−\mathrm{2}}}=\mathrm{1} \\ $$$${let}\:{t}=\sqrt[{\mathrm{3}}]{\frac{{x}}{{x}−\mathrm{2}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{t}^{\mathrm{3}} }{{t}^{\mathrm{3}} −\mathrm{1}} \\ $$$$\Rightarrow{x}−\mathrm{1}=\frac{{t}^{\mathrm{3}} +\mathrm{1}}{{t}^{\mathrm{3}} −\mathrm{1}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{\frac{{t}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} }}−\mathrm{1}\right){t}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{t}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}}}={t}+\mathrm{1} \\ $$$$\Rightarrow{t}^{\mathrm{3}} +\mathrm{1}=\mathrm{2}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right) \\ $$$$\Rightarrow{t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{6}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{t}_{{i}} =−\mathrm{6} \\ $$$${i}.{e}.\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\sqrt[{\mathrm{3}}]{\frac{{x}_{{i}} }{{x}_{{i}} −\mathrm{2}}}=−\mathrm{6} \\ $$$${i}.{e}.\:\sqrt[{\mathrm{3}}]{\frac{{a}}{{a}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}}=−\mathrm{6} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Jul/21
∨ ∩ ∣^(•) ⊂∈ $ir !
$$\vee\:\cap\:\overset{\bullet} {\shortmid}\:\subset\in\:\:\$\mathrm{ir}\:! \\ $$
Commented by Tawa11 last updated on 23/Jul/21
Sir mrW please tag the general symmetry polynomial you did sometimes ago. like, x + y + z = 1 x^2 + y^2 + z^2 = 2 x^3 + y^3 + z^3 = 3 then, x^5 + y^5 + z^5 = ?? Just tag the one you did sometimes ago. Thanks sir. God bless you.
$$\mathrm{Sir}\:\mathrm{mrW}\:\mathrm{please}\:\mathrm{tag}\:\mathrm{the}\:\mathrm{general}\:\mathrm{symmetry}\:\mathrm{polynomial}\:\mathrm{you}\:\mathrm{did} \\ $$$$\mathrm{sometimes}\:\mathrm{ago}. \\ $$$$\mathrm{like}, \\ $$$$\:\:\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{3} \\ $$$$\mathrm{then},\:\:\:\:\mathrm{x}^{\mathrm{5}} \:\:+\:\:\mathrm{y}^{\mathrm{5}} \:\:+\:\:\mathrm{z}^{\mathrm{5}} \:\:=\:\:?? \\ $$$$ \\ $$$$\mathrm{Just}\:\:\mathrm{tag}\:\mathrm{the}\:\mathrm{one}\:\mathrm{you}\:\mathrm{did}\:\mathrm{sometimes}\:\mathrm{ago}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by Tawa11 last updated on 23/Jul/21
Seen sir. Thanks.
$$\mathrm{Seen}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$
Commented by mnjuly1970 last updated on 23/Jul/21
thanks alot mr W
$$\:{thanks}\:{alot}\:{mr}\:{W} \\ $$
Answered by Rasheed.Sindhi last updated on 23/Jul/21
(x)^(1/3) +((x−2))^(1/3) =((x−1))^(1/3) ((a/(a−2)))^(1/3) +((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) =? Let x−1=y^3 ⇒x=y^3 +1 ((y^3 +1))^(1/3) +((y^3 −1))^(1/3) =y Cubing both sides: y^3 +1+y^3 −1+3(((y^3 +1)(y^3 −1))^(1/3) (y)=y^3 2y^3 +3y((y^6 −1))^(1/3) =y^3 y^3 +3y((y^6 −1))^(1/3) =0 y(y^2 +3((y^6 −1))^(1/3) )=0 y=0 ∣ y^2 +3((y^6 −1))^(1/3) =0 x−1=0 ∣ y^2 =−3((y^6 −1))^(1/3) x=1 ∣ y^6 =−27(y^6 −1) 28y^6 =27⇒(y^3 )^2 =((27)/(28)) (x−1)^2 =((27)/(28)) 28x^2 −56x+1=0 x=((56±(√(56^2 −4(28))))/(56)) x=((56±12(√(21)))/(56))=((14±3(√(21)))/(14)) a=1,b=((14+3(√(21)))/(14)),c=((14−3(√(21)))/(14)) A=((a/(a−2)))^(1/3) +((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) A=((1/(1−2)))^(1/3) +((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) A=−1+((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) (A+1)^3 =(((b/(b−2)))^(1/3) +((c/(c−2)))^(1/3) )^3 =(b/(b−2))+(c/(c−2))+3(((bc)/(bc−2(b+c)+4)))^(1/3) (A+1) =((bc−2b+bc−2c)/(bc−2(b+c)+4))+3(((bc)/(bc−2(b+c)+4)))^(1/3) (A+1) =((2bc−2(b+c))/(bc−2(b+c)+4))+3(((bc)/(bc−2(b+c)+4)))^(1/3) (A+1) bc=(((14+3(√(21)))/(14)))(((14−3(√(21)))/(14)))=((196−9(21))/(196)) =1/28 b+c=((14+3(√(21)))/(14))+((14−3(√(21)))/(14))=2 (A+1)^3 =((2(1/28)−2(2))/((1/28)−2(2)+4))+3(((1/28)/((1/28)−2(2)+4)))^(1/3) (A+1) (A+1)^3 =(((1/(14))−4)/(1/(28)))+3(((1/28)/((1/28))))^(1/3) (A+1) (A+1)^3 =(((−55)/(14))/(1/(28)))+3(((1/28)/((1/28))))^(1/3) (A+1) =−110+3A+3 (A+1)^3 −3A+107=0 A^3 +1+3A(A+1)−3A+107=0o A^3 +3A^2 +3A−3A+108=0 A^3 +3A^2 +108=0 (A+6)(A^2 −3A+18)=0 A=−6, ((3±(√(9−72)))/2)=((3±3i(√7))/2)(Rejected because A is obviously real)
$$\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}=\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{{a}}{{a}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}}=? \\ $$$${Let}\:{x}−\mathrm{1}={y}^{\mathrm{3}} \Rightarrow{x}={y}^{\mathrm{3}} +\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{y}^{\mathrm{3}} +\mathrm{1}}+\sqrt[{\mathrm{3}}]{{y}^{\mathrm{3}} −\mathrm{1}}={y} \\ $$$${Cubing}\:{both}\:{sides}: \\ $$$${y}^{\mathrm{3}} +\mathrm{1}+{y}^{\mathrm{3}} −\mathrm{1}+\mathrm{3}\sqrt[{\mathrm{3}}]{\left({y}^{\mathrm{3}} +\mathrm{1}\right)\left({y}^{\mathrm{3}} −\mathrm{1}\right.}\left({y}\right)={y}^{\mathrm{3}} \\ $$$$\mathrm{2}{y}^{\mathrm{3}} +\mathrm{3}{y}\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} −\mathrm{1}}={y}^{\mathrm{3}} \\ $$$${y}^{\mathrm{3}} +\mathrm{3}{y}\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} −\mathrm{1}}=\mathrm{0} \\ $$$${y}\left({y}^{\mathrm{2}} +\mathrm{3}\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} −\mathrm{1}}\right)=\mathrm{0} \\ $$$${y}=\mathrm{0}\:\mid\:{y}^{\mathrm{2}} +\mathrm{3}\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} −\mathrm{1}}=\mathrm{0} \\ $$$${x}−\mathrm{1}=\mathrm{0}\:\mid\:{y}^{\mathrm{2}} =−\mathrm{3}\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} −\mathrm{1}} \\ $$$$\:{x}=\mathrm{1}\:\mid\:\:{y}^{\mathrm{6}} =−\mathrm{27}\left({y}^{\mathrm{6}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{28}{y}^{\mathrm{6}} =\mathrm{27}\Rightarrow\left({y}^{\mathrm{3}} \right)^{\mathrm{2}} =\frac{\mathrm{27}}{\mathrm{28}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{27}}{\mathrm{28}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{28}{x}^{\mathrm{2}} −\mathrm{56}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}=\frac{\mathrm{56}\pm\sqrt{\mathrm{56}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{28}\right)}}{\mathrm{56}} \\ $$$$\:\:\:{x}=\frac{\mathrm{56}\pm\mathrm{12}\sqrt{\mathrm{21}}}{\mathrm{56}}=\frac{\mathrm{14}\pm\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}} \\ $$$${a}=\mathrm{1},{b}=\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}},{c}=\frac{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}} \\ $$$${A}=\sqrt[{\mathrm{3}}]{\frac{{a}}{{a}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}} \\ $$$${A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}} \\ $$$${A}=−\mathrm{1}+\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}} \\ $$$$\left({A}+\mathrm{1}\right)^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\frac{{b}}{{b}−\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{c}−\mathrm{2}}}\right)^{\mathrm{3}} \\ $$$$\:=\frac{{b}}{{b}−\mathrm{2}}+\frac{{c}}{{c}−\mathrm{2}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{bc}}{{bc}−\mathrm{2}\left({b}+{c}\right)+\mathrm{4}}}\left({A}+\mathrm{1}\right) \\ $$$$\:=\frac{{bc}−\mathrm{2}{b}+{bc}−\mathrm{2}{c}}{{bc}−\mathrm{2}\left({b}+{c}\right)+\mathrm{4}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{bc}}{{bc}−\mathrm{2}\left({b}+{c}\right)+\mathrm{4}}}\left({A}+\mathrm{1}\right) \\ $$$$\:=\frac{\mathrm{2}{bc}−\mathrm{2}\left({b}+{c}\right)}{{bc}−\mathrm{2}\left({b}+{c}\right)+\mathrm{4}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{bc}}{{bc}−\mathrm{2}\left({b}+{c}\right)+\mathrm{4}}}\left({A}+\mathrm{1}\right) \\ $$$${bc}=\left(\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}\right)\left(\frac{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}\right)=\frac{\mathrm{196}−\mathrm{9}\left(\mathrm{21}\right)}{\mathrm{196}} \\ $$$$\:\:\:\:=\mathrm{1}/\mathrm{28} \\ $$$${b}+{c}=\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}+\frac{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}=\mathrm{2} \\ $$$$\left({A}+\mathrm{1}\right)^{\mathrm{3}} =\frac{\mathrm{2}\left(\mathrm{1}/\mathrm{28}\right)−\mathrm{2}\left(\mathrm{2}\right)}{\left(\mathrm{1}/\mathrm{28}\right)−\mathrm{2}\left(\mathrm{2}\right)+\mathrm{4}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}/\mathrm{28}}{\left(\mathrm{1}/\mathrm{28}\right)−\mathrm{2}\left(\mathrm{2}\right)+\mathrm{4}}}\left({A}+\mathrm{1}\right) \\ $$$$\left({A}+\mathrm{1}\right)^{\mathrm{3}} =\frac{\frac{\mathrm{1}}{\mathrm{14}}−\mathrm{4}}{\frac{\mathrm{1}}{\mathrm{28}}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}/\mathrm{28}}{\left(\mathrm{1}/\mathrm{28}\right)}}\left({A}+\mathrm{1}\right) \\ $$$$\left({A}+\mathrm{1}\right)^{\mathrm{3}} =\frac{\frac{−\mathrm{55}}{\mathrm{14}}}{\frac{\mathrm{1}}{\mathrm{28}}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}/\mathrm{28}}{\left(\mathrm{1}/\mathrm{28}\right)}}\left({A}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:=−\mathrm{110}+\mathrm{3}{A}+\mathrm{3} \\ $$$$\left({A}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}{A}+\mathrm{107}=\mathrm{0} \\ $$$${A}^{\mathrm{3}} +\mathrm{1}+\mathrm{3}{A}\left({A}+\mathrm{1}\right)−\mathrm{3}{A}+\mathrm{107}=\mathrm{0}{o} \\ $$$${A}^{\mathrm{3}} +\mathrm{3}{A}^{\mathrm{2}} +\mathrm{3}{A}−\mathrm{3}{A}+\mathrm{108}=\mathrm{0} \\ $$$${A}^{\mathrm{3}} +\mathrm{3}{A}^{\mathrm{2}} +\mathrm{108}=\mathrm{0} \\ $$$$\left({A}+\mathrm{6}\right)\left({A}^{\mathrm{2}} −\mathrm{3}{A}+\mathrm{18}\right)=\mathrm{0} \\ $$$${A}=−\mathrm{6},\:\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{72}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\mathrm{3}{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\left({Rejected}\right. \\ $$$$\left.{because}\:{A}\:{is}\:{obviously}\:{real}\right) \\ $$
Commented by mr W last updated on 23/Jul/21
good too!
$${good}\:{too}! \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jul/21
An alternate although some what more time consuming yet not bad for practising.
$$\mathcal{A}{n}\:{alternate} \\ $$$${although} \\ $$$${some}\:{what}\:{more}\:{time}\:{consuming} \\ $$$${yet} \\ $$$${not}\:{bad}\:{for}\:{practising}. \\ $$
Commented by Rasheed.Sindhi last updated on 23/Jul/21
Thank You sir mr W!
$$\mathcal{T}{hank}\:\mathcal{Y}{ou} \\ $$$$\boldsymbol{{sir}}\:{mr}\:{W}! \\ $$

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