If 3x+(1/(2x))=6 find 8x^3 +(1/(27x^3 ))


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Question Number 119713 by som(math1967) last updated on 26/Oct/20

$If 3 x + \frac{1}{2 x} = 6 find {8 x}^{3} + \frac{1}{{27 x}^{3}}$
Commented by bemath last updated on 26/Oct/20

$s a n t u y y$
Commented by Jamshidbek2311 last updated on 26/Oct/20

$3 x + \frac{1}{2 x} = 6 ∣ \times \frac{2}{3} \Rightarrow 2 x + \frac{1}{3 x} = 4$ $8 x^{3} + \frac{1}{27 x^{3}} = {(2 x + \frac{1}{3 x})}^{3} - 3 \times 2 x \times \frac{1}{3 x} (2 x + \frac{1}{3 x}) =$ $= 4^{3} - 2 \times 4 = 64 - 8 = 56$
Commented by som(math1967) last updated on 26/Oct/20

$Thanks$
	Answered by Dwaipayan Shikari last updated on 26/Oct/20

	$8 x^{3} + \frac{1}{27 x^{3}} = {(2 x + \frac{1}{3 x})}^{3} - 2 (2 x + \frac{1}{3 x}) = 4^{3} - 2.4 = 56$ $\frac{6 x^{2} + 1}{2 x} = 6$ $\Rightarrow \frac{6 x^{2} + 1}{3 x} = 4 \Rightarrow 2 x + \frac{1}{3 x} = 4$
	Commented by som(math1967) last updated on 26/Oct/20

	$Very nice thank you$
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