Question

If 3x+1/3x=3, find : 27x^3+1/27x^3.

Answer 1

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Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{3x + \frac{1}{3x} = 3 }$

$\bf \underline{To\: find-}$

$\sf{ the \:value \: of : \: {27x}^{3} + \frac{1}{ {27x}^{3} } = \: ?}$

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\sf{3x + \frac{1}{3x} = 3 }$

$\textsf{Cubing on both sides, we get}$

$\sf{ \bigg(3x + \frac{1}{3x} \bigg)^{3} = (3 {)}^{3} }$

$\textsf{★Now, comparing this expression with (a+b)³, we get}$

$\sf{ \: \: \: \: a = 3x \: and \: b = \frac{1}{3x}. }$

$★\textsf{Using identity (a+b)³=a³+b³+3ab(a+b),we get}$

$\sf{ \bigg(3x + \frac{1}{3x} \bigg)^{3} = (3 {)}^{3} }$

$\sf{ \implies \: (3x {)}^{3} + \bigg(\frac{1}{3x} \bigg) ^{3} + 3(3x) \bigg( \frac{1}{3x} \bigg) \bigg(3x + \frac{1}{3x} \bigg) = 27}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} }+ 3(3x) \bigg( \frac{1}{3x} \bigg) \bigg(3x + \frac{1}{3x} \bigg) = 27}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} }+ 3( \cancel{3x}) \bigg( \frac{1}{ \cancel{3x}} \bigg) \bigg(3x + \frac{1}{3x} \bigg) = 27}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} }+ 3 \bigg(3x + \frac{1}{3x} \bigg) = 27}$

$\textsf{★Since 3x + \frac{1}{3x} = 3 (Given)}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} }+ 3 (3)= 27}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} } + 9= 27}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} } = 27 - 9}$

$\sf{ \implies \: 27x ^{3} + \frac{1}{27x ^{3} } = 18}$

$\bf \underline{Answer-}$

$\bf \underline{Hence, the\: value\:of : \: 27x ^{3} + \frac{1}{27x} ^{3} \: is \:18.}$