Is Quadratic Formula Relevant Here?

$\Rightarrow 3x+\dfrac{1}{2x}=3$

Cubing both sides.

$(3x)^3+\left( \dfrac{1}{2x} \right)^3+3×3x×\dfrac{1}{2x}\left(3x+\dfrac{1}{2x}\right)=27$

$27x^3+\dfrac{1}{8x^3}+\dfrac{27}{2}=27$

$27x^3+\dfrac{1}{8x^3}=\dfrac{27}{2}$

Multiplying $\dfrac{8}{27}$ both sides.

$\left(27x^3×\dfrac{8}{27}\right)+\left(\dfrac{1}{8x^3}×\dfrac{8}{27}\right)=\dfrac{27}{2}×\dfrac{8}{27}$

$\Rightarrow 8x^3+\dfrac{1}{27x^3}=\boxed{4}$

3 $x$ + $\frac{1}{2x}$ = 3 $=>$ 2 $x$ + $\frac{1}{3x}$ = 2 Cube the equation: 8 $x^3$ + $\frac{1}{27x^3}$ + 3 x $2x$ x $\frac{1}{3x}$ x ( 2 $x$ + $\frac{1}{3x}$ ) = 8 => 8 $x^3$ + $\frac{1}{27x^3}$ = 8 -4 = 4

3x + $\frac{1}{2x}$ =3 , multiply both sides by 2x and then divide by 3x to obtain 2x+ $\frac{1}{3x}$ =2

now take cube on both sides

using formula : $(a+b)^3 =a^3 +b^3 + 3ab(a+b)$

=> $(2x)^3$ + $(\frac{1}{3x})^3$ + $3 (2x)$ ( $\frac{1}{3x}$ )[ $2x$ + $\frac{1}{3x}$ ]= $2^3$

(bcoz $2x$ + $\frac{1}{3x}$ =2

$(2x)^3$ + $(\frac{1}{3x})^3$ + $3 (2x)$ ( $\frac{1}{3x}$ )[2]= $2^3$

now after cancelling the same terms we have

$(2x)^3$ + $(\frac{1}{3x})^3$ +2[2]= $2^3$

$8x^3$ + $\frac{1}{27x^3}$ +4=8

=> $8x^3$ + $\frac{1}{27x^3}$ =8-4 => 4 ANS....

Multiply the given equation with $\frac{2}{3}$ and cube the obtained expression and simplify further to obtain the result as $\boxed{4}$

$8x^3 + \dfrac1{27x^3}$

$= (2x + \frac{1}{3x})(4x^2 - \frac{2x}{3x} + \frac{1}{9x^2})$

$= (2x + \frac{1}{3x})(4x^2 - \frac{2}{3} + \frac{1}{9x^2})$

We know that $a^2 - ab + b^2 = (a+b)^2 - 3ab$ , hence the expression above is:

$= (2x + \frac{1}{3x})((2x + \frac{1}{3x})^2 - 3(\frac{2}{3})) = (2x + \frac{1}{3x})[(2x + \frac{1}{3x})^2 - 2]$

We need to find $(2x + \frac{1}{3x})$ :

From the 1st equation that was given ( $3x + \dfrac1{2x} = 3$ )

$\frac{2}{3} \times (3x + \dfrac1{2x}) = 3 \times \frac{2}{3}$

$\boxed{ 2x + \frac{1}{3x} = 2 }$

Therefore:

$8x^3 + \dfrac1{27x^3} = (2x + \frac{1}{3x})[(2x + \frac{1}{3x})^2 - 2] = (2)(2^2 - 2) = \boxed{4}$

Here my way frnds first we rearrange the term that is given in the question I.e 3x +1/2x =3 3(x-1)= -1/2x 2(x-1)=-1/3x 2x +1/3x = 2 8x^3+ 1/27x^3 in to factorize by using a^3 + b ^3 formula nd plug the value obtain from above to find the answer .

We have that $3x +\dfrac{1}{2x} = 3 \implies 2x = \dfrac{6x-1}{3x}$ And we need to evaluate $E = 8x^3 +\dfrac{1}{27x^3} = \left(2x+\dfrac{1}{3x}\right)\left(\left(2x+\dfrac{1}{3x}\right)^2 -2 \dfrac{2x}{3x} - \dfrac{2x}{3x}\right)\cdots (1)$ Now plugging $2x = \dfrac{6x-1}{3x}$ in the 1st equation we get $E = \left(\dfrac{6x-1}{3x}+\dfrac{1}{3x}\right)\left(\left(\dfrac{6x-1}{3x} +\dfrac{1}{3x}\right)^2 -\dfrac{4}{3} -\dfrac{2}{3}\right)\\\boxed{ E = 2\,(2^2 - 2) =\boxed{4}}$

This question just requires noticing that $2x+\dfrac{1}{3x}=\dfrac{2}{3}(3x+\dfrac{1}{2x})$

Thus, we get that $2x+\dfrac{1}{3x}=\dfrac{2}{3}.3=2$

And then we use the fact that

$8x^{3}+\dfrac{1}{27x^{3}}=(2x+\dfrac{1}{3x})^{3}-3(2x)(\dfrac{1}{3x})(2x+\dfrac{1}{3x})=(2x+\dfrac{1}{3x})^{3}-2(2x+\dfrac{1}{3x})$

And thus we get

$8x^{3}+\dfrac{1}{27x^{3}}=2^{3}-2(2)=8-4=\boxed {4}$