Q-If-a-b-are-positive-numbers-and-a-1-6a-2-1-3-b-1-6b-2-1-3-then-find-the-value-of-a-b-Compile

Question Number 152939 by mnjuly1970 last updated on 03/Sep/21

Q : If a, b are positive numbers and

{\begin{cases} a = 1 + \sqrt[3]{6 a - 2} \\ b = 1 + \sqrt[3]{6 b - 2} \end{cases}

then find the value of, a . b = ?

\dots Compiled by m . n : (E l e m e n t a r y o l y m p i a d) .

Answered by mr W last updated on 03/Sep/21

a, b a r e r o o t s o f

x = 1 + \sqrt[3]{6 x - 2}

{(x - 1)}^{3} = 6 x - 2

x^{3} - 3 x^{2} - 3 x + 1 = 0

(x + 1) (x^{2} - 4 x + 1) = 0

p o s i t i v e r o o t s a, b a r e r o o t s o f x^{2} - 4 x + 1 = 0

\Rightarrow a + b = 4, a b = 1

Commented by mnjuly1970 last updated on 03/Sep/21

b r a v o s i r W . g r a t e f u l . .

Commented by mr W last updated on 03/Sep/21

t h a n k s f o r t h e n i c e q u e s t i o n s s i r!

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