{ (((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=?


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Question Number 161484 by bobhans last updated on 18/Dec/21
${ (((1/a)+(1/b)=9)),((((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) )))=18)) :} 8a+4b=?$
${\begin{cases} \frac{1}{a} + \frac{1}{b} = 9 \\ (\frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}}) (1 + \frac{1}{\sqrt[3]{a}}) (1 + \frac{1}{\sqrt[3]{b}}) = 18 \end{cases}$ $8 a + 4 b = ?$
	Answered by mr W last updated on 18/Dec/21

	$l e t A = \frac{1}{\sqrt[3]{a}}, B = \frac{1}{\sqrt[3]{b}}$ $l e t p = A + B, q = A B$ $A^{3} + B^{3} = 9$ $(A + B) [{(A + B)}^{2} - 3 A B] = 9$ $p^{3} - 3 p q = 9 . . . (i)$ $(A + B) (1 + A) (1 + B) = 18$ $(A + B) (A + B + 1 + A B) = 18$ $p^{2} + p + p q = 18 . . . (i i)$ $(i) + 3 \times (i i) :$ $p^{3} + 3 p^{2} + 3 p = 63$ ${(p + 1)}^{3} = 64$ $\Rightarrow p + 1 = 4$ $\Rightarrow p = 3 = A + B$ $3^{3} - 3 \times 3 q = 9$ $\Rightarrow q = 2 = A B$ $A, B a r e r o o t s o f x^{2} - 3 x + 2 = 0$ $\Rightarrow (A, B) = (x_{1}, x_{2}) = (1, 2)$ $a = \frac{1}{A^{3}}, b = \frac{1}{B^{3}}$ $\Rightarrow (a, b) = (1, \frac{1}{8})$ $8 a + 4 b = 8 \times 1 + 4 \times \frac{1}{8} = \frac{17}{2} o r$ $8 a + 4 b = 8 \times \frac{1}{8} + 4 \times 1 = 5$
	Answered by blackmamba last updated on 18/Dec/21
	$(Q) { (((1/a)+(1/b)=9)),((((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) )))= 18)) :} ⇔ (1+(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))^3 =1+(1/a)+(1/b)+3((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))(1+(1/( (a)^(1/3) )))(1+(1/( (b)^(1/3) ))) ⇔ (1+(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))^3 = 1+9+3(18)=64 ⇔(1/( (a)^(1/3) ))+(1/( (b)^(1/3) )) = 3 ; (1/a)+(1/b)+3((1/( ((ab))^(1/3) )))((1/( (a)^(1/3) ))+(1/( (b)^(1/3) )))=27 ⇒9+3(3)((1/( ((ab))^(1/3) )))=27 ; ab = (1/8) ⇒ { (((1/a)+(1/b)=((a+b)/(ab))= 9⇒a+b=(9/8))),((ab=(1/8))) :} ⇒a((9/8)−a)=(1/8); 9a−8a^2 =1 ⇒8a^2 −9a+1=0 ⇒(8a−1)(a−1)=0 ⇒ { ((a=1 ∧ b=(1/8))),((a=(1/8) ∧ b=1)) :} ⇒8a+4b = { ((8+(1/2)=((17)/2) ; or)),((1+4 = 5 )) :}$
	$(Q) {\begin{cases} \frac{1}{a} + \frac{1}{b} = 9 \\ (\frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}}) (1 + \frac{1}{\sqrt[3]{a}}) (1 + \frac{1}{\sqrt[3]{b}}) = 18 \end{cases}$ $\Leftrightarrow {(1 + \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}})}^{3} = 1 + \frac{1}{a} + \frac{1}{b} + 3 (\frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}}) (1 + \frac{1}{\sqrt[3]{a}}) (1 + \frac{1}{\sqrt[3]{b}})$ $\Leftrightarrow {(1 + \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}})}^{3} = 1 + 9 + 3 (18) = 64$ $\Leftrightarrow \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}} = 3; \frac{1}{a} + \frac{1}{b} + 3 (\frac{1}{\sqrt[3]{a b}}) (\frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}}) = 27$ $\Rightarrow 9 + 3 (3) (\frac{1}{\sqrt[3]{a b}}) = 27; a b = \frac{1}{8}$ $\Rightarrow {\begin{cases} \frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = 9 \Rightarrow a + b = \frac{9}{8} \\ a b = \frac{1}{8} \end{cases}$ $\Rightarrow a (\frac{9}{8} - a) = \frac{1}{8}; 9 a - 8 a^{2} = 1$ $\Rightarrow 8 a^{2} - 9 a + 1 = 0$ $\Rightarrow (8 a - 1) (a - 1) = 0 \Rightarrow {\begin{cases} a = 1 \land b = \frac{1}{8} \\ a = \frac{1}{8} \land b = 1 \end{cases}$ $\Rightarrow 8 a + 4 b = {\begin{cases} 8 + \frac{1}{2} = \frac{17}{2}; o r \\ 1 + 4 = 5 \end{cases}$
	Commented by Tawa11 last updated on 18/Dec/21

	$Great sir .$
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