Question

If x>0 and x²+1/9x²=25/36, find :x³+1/27x³


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Answer 1

Given

• ${ {x}^{2} + \cfrac{1}{9 {x}^{2} } = \cfrac{25}{36} \: \: \: \: \: \: \: - - - - - (1)}$

To Find

• ${x}^{2} + \cfrac{1}{27{x}^{3} }$

Formula used

$\boxed{\bf{a²+b²=(a+b)²-2ab}}$

$\boxed{\bf{a³+b³=(a+b)(a²+b²-ab)}}$

Solution

$\sf {x}^{2} + \cfrac{1}{9 {x}^{2} } = \cfrac{25}{36}$

$\sf {x}^{2} + \cfrac{1}{ {3x}^{2} } = \cfrac{25}{36}$

$\boxed{\bf{a²+b²=(a+b)²-2ab}}$

$\sf \left(x + \dfrac{1}{3c} \right)^{2} - 2 \times x \times \dfrac{1}{3x} = \cfrac{25}{36}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} - \dfrac{2}{3} = \cfrac{25}{36}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{25}{36} + \cfrac{2}{3}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{25 + 24}{36}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{49}{36}$

$\sf \left(x + \dfrac{1}{3x} \right) = \sqrt{ \cfrac{ {7}^{2} }{ {6}^{2} }}$

$\sf \left(x + \dfrac{1}{3x} \right) = \cfrac{7}{6} \: \: \: \: \: \: \: - - - - (2)$

Now

$\\\sf {x}^{2} + \cfrac{1}{27{x}^{3} }$

$\sf {x}^{2} + \cfrac{1}{{(3x)}^{3} }$

$\boxed{\bf{a³+b³=(a+b)(a²+b²-ab)}}$

$\sf {{x}^{2} + \cfrac{1}{(3{x})^{3} } = (x + \cfrac{1}{3x} )( {x}^{2} + \cfrac{1}{ {3x}^{2} } - 2 \times x \times \cfrac{1}{3x})}$

$\sf {{x}^{2} + \cfrac{1}{(3{x})^{3} } = ( \cfrac{7}{6} )( \cfrac{25}{36} - \cfrac{1}{3})}$

${\sf {x}^{2} + \cfrac{1}{(3{x})^{3} } = ( \cfrac{7}{6} )( \cfrac{25 - 24}{36} )}$

${\sf {x}^{2} + \cfrac{1}{(3{x})^{3} } = \cfrac{7}{6} \times \cfrac{1} {36} }$

$\bf {{x}^{2} + \cfrac{1}{27{x}^{3} } = \ \cfrac{7} {216} }$

Answer 2

${ {x}^{2} + \cfrac{1}{9 {x}^{2} } = \cfrac{25}{36} \: \: \: \: \: \: \: - - - - - (1)}$

${x}^{2} + \cfrac{1}{27{x}^{3} }$

$\huge\boxed{\bf{a²+b²=(a+b)²-2ab}}$

$\huge\boxed{\bf{a³+b³=(a+b)(a²+b²-ab)}}$

$\huge\sf {x}^{2} + \cfrac{1}{9 {x}^{2} } = \cfrac{25}{36}x$

$\huge\sf {x}^{2} + \cfrac{1}{ {3x}^{2} } = \cfrac{25}{36}$

$\huge\boxed{\bf{a²+b²=(a+b)²-2ab}}$

$\sf \huge\left(x + \dfrac{1}{3c} \right)^{2} - 2 \times x \times \dfrac{1}{3x} = \cfrac{25}{36}(x+$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} - \dfrac{2}{3} = \cfrac{25}{36}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{25}{36} + \cfrac{2}{3}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{25 + 24}{36}$

$\sf \left(x + \dfrac{1}{3x} \right)^{2} = \cfrac{49}{36}$

$\sf \left(x + \dfrac{1}{3x} \right) = \sqrt{ \cfrac{ {7}^{2} }{ {6}^{2} }}$

$\begin{gathered}\sf \left(x + \dfrac{1}{3x} \right) = \cfrac{7}{6} \: \: \: \: \: \: \: - - - - (2)\\\end{gathered}$

$\begin{gathered}\\\sf {x}^{2} + \cfrac{1}{27{x}^{3} } \end{gathered}$

$\sf {x}^{2} + \cfrac{1}{{(3x)}^{3} }$

$\huge\boxed{\bf{a³+b³=(a+b)(a²+b²-ab)}}$

$\sf {{x}^{2} + \cfrac{1}{(3{x})^{3} } = (x + \cfrac{1}{3x} )( {x}^{2} + \cfrac{1}{ {3x}^{2} } - 2 \times x \times \cfrac{1}{3x})}$

$\sf {{x}^{2} + \cfrac{1}{(3{x})^{3} } = ( \cfrac{7}{6} )( \cfrac{25}{36} - \cfrac{1}{3})}$

${\sf {x}^{2} + \cfrac{1}{(3{x})^{3} } = ( \cfrac{7}{6} )( \cfrac{25 - 24}{36} )}$

${\sf {x}^{2} + \cfrac{1}{(3{x})^{3} } = \cfrac{7}{6} \times \cfrac{1} {36} }$

$\huge \begin{gathered}\bf {{x}^{2} + \cfrac{1}{27{x}^{3} } = \ \cfrac{7} {216} }\\\end{gathered}$