Question

If x is a positive number such that the ratio 3x to
$\sqrt[3]{x}$
is equal to 27:1 then find then find the value of
$x ^{3 } ratio \: 27 \sqrt[3]{x {}^{2} }$

​

Answer 1

Answer:

please mark me brainliest

Answer 2

Given:$\[3x:\sqrt[3]{x} = 27:1\]$.

To find: the value of $\[{x^3}:27\sqrt[3]{{{x^2}}}\]$.

Solution:

Observe that from the question and find the value of x.

$\[3x:\sqrt[3]{x} = 27:1\]$

$\[ \Rightarrow \frac{{3x}}{{\sqrt[3]{x}}} = \frac{{27}}{1}\]$

$\[\begin{array}{l} \Rightarrow 3x \times 1 = 27 \times \sqrt[3]{x}\\ \Rightarrow 3x = 27\sqrt[3]{x}\\ \Rightarrow 27{x^3} = {\left( {27} \right)^3}x\end{array}\]$

$\[\begin{array}{l} \Rightarrow {x^2} = {27^2}\\ \Rightarrow x = \pm 27\end{array}\]$

Find the value of$\[{x^3}:27\sqrt[3]{{{x^2}}}\]$.

$\[\frac{{{x^3}}}{{27\sqrt[3]{{{x^2}}}}} = \frac{{\left( {{{27}^3}} \right)}}{{27 \times \sqrt[3]{{{{27}^2}}}}}\]$

         $\[ = \frac{{{{27}^2}}}{{\sqrt[3]{{{{27}^2}}}}}\]$

         $\[ = \frac{{729}}{9}\]$

         $\[ = \frac{{81}}{1}\]$

Hence, $\[{x^3}:27\sqrt[3]{{{x^2}}} = 81:1\]$.