Question

x upon 3 -3 upon x the whole 3

Answer 1

Answer:

$\left(\frac{x}{3}-\frac{3}{x}\right)^{3}\\= \frac{x^{3}}{27}-x+\frac{9}{x}-\frac{27}{x^{3}}$

Step-by-step explanation:

$\left(\frac{x}{3}-\frac{3}{x}\right)^{3}$

$=\left(\frac{x}{3}\right)^{3}-3\times \left(\frac{x}{3}\right)^{2}\times \left(\frac{3}{x}\right)+ 3\times \left(\frac{x}{3}\right)\times \left(\frac{3}{x}\right)^{2} - \left(\frac{3}{x}\right)^{3}$

/* By algebraic identity:

(a-b)³ = a³-3a²b+3ab²-b³ */

$= \frac{x^{3}}{27}-x+\frac{9}{x}-\frac{27}{x^{3}}$

Therefore,

$\left(\frac{x}{3}-\frac{3}{x}\right)^{3}\\= \frac{x^{3}}{27}-x+\frac{9}{x}-\frac{27}{x^{3}}$

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Answer 2

The value of $(\dfrac{x}{3}-\dfrac{3}{x})^3$ is  $\dfrac{x^3}{27}-x+\dfrac{9}{x}-\dfrac{27}{x^3}$.

Step-by-step explanation:

We need to solve the given expression i.e.

$(\dfrac{x}{3}-\dfrac{3}{x})^3$

It can be done using identity : $(a-b)^3=a^3 - 3a^2b + 3ab^2 - b^3$

We have,

$a=\dfrac{x}{3}\\\\b=\dfrac{3}{x}$

So,

$(\dfrac{x}{3}-\dfrac{3}{x})^3=(\dfrac{x}{3})^3-3\times (\dfrac{x}{3})^2\times \dfrac{3}{x}+3\times \dfrac{x}{3}\times (\dfrac{3}{x})^2-(\dfrac{3}{x})^3\\\\=\dfrac{x^3}{27}-x+\dfrac{9}{x}-\dfrac{27}{x^3}$

So, the value of $(\dfrac{x}{3}-\dfrac{3}{x})^3$ is  $\dfrac{x^3}{27}-x+\dfrac{9}{x}-\dfrac{27}{x^3}$.

Learn more,

Algebraic identity

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