Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 161484 by bobhans last updated on 18/Dec/21

{ (((1/a)+(1/b)=9)),((((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) )))=18)) :} 8a+4b=?

$$\:\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\mathrm{9}}\\{\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)=\mathrm{18}}\end{cases} \\ $$$$\:\:\:\mathrm{8a}+\mathrm{4b}=? \\ $$

Answered by mr W last updated on 18/Dec/21

let A=(1/( (a)^(1/3) )), B=(1/( (b)^(1/3) )) let p=A+B, q=AB A^3 +B^3 =9 (A+B)[(A+B)^2 −3AB]=9 p^3 −3pq=9 ...(i) (A+B)(1+A)(1+B)=18 (A+B)(A+B+1+AB)=18 p^2 +p+pq=18 ...(ii) (i)+3×(ii): p^3 +3p^2 +3p=63 (p+1)^3 =64 ⇒p+1=4 ⇒p=3=A+B 3^3 −3×3q=9 ⇒q=2=AB A, B are roots of x^2 −3x+2=0 ⇒(A,B)=(x_1 ,x_2 )=(1,2) a=(1/A^3 ), b=(1/B^3 ) ⇒(a,b)=(1,(1/8)) 8a+4b=8×1+4×(1/8)=((17)/2) or 8a+4b=8×(1/8)+4×1=5

$${let}\:{A}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}},\:{B}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}} \\ $$$${let}\:{p}={A}+{B},\:{q}={AB} \\ $$$$ \\ $$$${A}^{\mathrm{3}} +{B}^{\mathrm{3}} =\mathrm{9} \\ $$$$\left({A}+{B}\right)\left[\left({A}+{B}\right)^{\mathrm{2}} −\mathrm{3}{AB}\right]=\mathrm{9} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{pq}=\mathrm{9}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\left({A}+{B}\right)\left(\mathrm{1}+{A}\right)\left(\mathrm{1}+{B}\right)=\mathrm{18} \\ $$$$\left({A}+{B}\right)\left({A}+{B}+\mathrm{1}+{AB}\right)=\mathrm{18} \\ $$$${p}^{\mathrm{2}} +{p}+{pq}=\mathrm{18}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\mathrm{3}×\left({ii}\right): \\ $$$${p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{p}=\mathrm{63} \\ $$$$\left({p}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{64} \\ $$$$\Rightarrow{p}+\mathrm{1}=\mathrm{4} \\ $$$$\Rightarrow{p}=\mathrm{3}={A}+{B} \\ $$$$\mathrm{3}^{\mathrm{3}} −\mathrm{3}×\mathrm{3}{q}=\mathrm{9} \\ $$$$\Rightarrow{q}=\mathrm{2}={AB} \\ $$$${A},\:{B}\:{are}\:{roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({A},{B}\right)=\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \right)=\left(\mathrm{1},\mathrm{2}\right) \\ $$$${a}=\frac{\mathrm{1}}{{A}^{\mathrm{3}} },\:{b}=\frac{\mathrm{1}}{{B}^{\mathrm{3}} } \\ $$$$\Rightarrow\left({a},{b}\right)=\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{8}×\mathrm{1}+\mathrm{4}×\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{17}}{\mathrm{2}}\:{or} \\ $$$$\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{8}×\frac{\mathrm{1}}{\mathrm{8}}+\mathrm{4}×\mathrm{1}=\mathrm{5} \\ $$

Answered by blackmamba last updated on 18/Dec/21

(Q) { (((1/a)+(1/b)=9)),((((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) )))= 18)) :} ⇔ (1+(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))^3 =1+(1/a)+(1/b)+3((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) ))) ⇔ (1+(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))^3 = 1+9+3(18)=64 ⇔(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )) = 3 ; (1/a)+(1/b)+3((1/( ((ab))^(1/3) )))((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))=27 ⇒9+3(3)((1/( ((ab))^(1/3) )))=27 ; ab = (1/8) ⇒ { (((1/a)+(1/b)=((a+b)/(ab))= 9⇒a+b=(9/8))),((ab=(1/8))) :} ⇒a((9/8)−a)=(1/8); 9a−8a^2 =1 ⇒8a^2 −9a+1=0 ⇒(8a−1)(a−1)=0 ⇒ { ((a=1 ∧ b=(1/8))),((a=(1/8) ∧ b=1)) :} ⇒8a+4b = { ((8+(1/2)=((17)/2) ; or)),((1+4 = 5 )) :}

$$\:\left({Q}\right)\:\begin{cases}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\mathrm{9}}\\{\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)=\:\mathrm{18}}\end{cases} \\ $$$$\:\Leftrightarrow\:\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)^{\mathrm{3}} =\mathrm{1}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right) \\ $$$$\:\Leftrightarrow\:\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)^{\mathrm{3}} =\:\mathrm{1}+\mathrm{9}+\mathrm{3}\left(\mathrm{18}\right)=\mathrm{64} \\ $$$$\:\Leftrightarrow\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\:=\:\mathrm{3}\:;\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ab}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)=\mathrm{27} \\ $$$$\Rightarrow\mathrm{9}+\mathrm{3}\left(\mathrm{3}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ab}}}\right)=\mathrm{27}\:;\:{ab}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\begin{cases}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{{a}+{b}}{{ab}}=\:\mathrm{9}\Rightarrow{a}+{b}=\frac{\mathrm{9}}{\mathrm{8}}}\\{{ab}=\frac{\mathrm{1}}{\mathrm{8}}}\end{cases} \\ $$$$\Rightarrow{a}\left(\frac{\mathrm{9}}{\mathrm{8}}−{a}\right)=\frac{\mathrm{1}}{\mathrm{8}};\:\mathrm{9}{a}−\mathrm{8}{a}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{8}{a}^{\mathrm{2}} −\mathrm{9}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{8}{a}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\begin{cases}{{a}=\mathrm{1}\:\wedge\:{b}=\frac{\mathrm{1}}{\mathrm{8}}}\\{{a}=\frac{\mathrm{1}}{\mathrm{8}}\:\wedge\:{b}=\mathrm{1}}\end{cases} \\ $$$$\:\Rightarrow\mathrm{8}{a}+\mathrm{4}{b}\:=\:\begin{cases}{\mathrm{8}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{17}}{\mathrm{2}}\:;\:{or}}\\{\mathrm{1}+\mathrm{4}\:=\:\mathrm{5}\:}\end{cases} \\ $$

Commented by Tawa11 last updated on 18/Dec/21

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com