Question

If $x=(-3)$, then find the value of $(x^2{)}^3\times {\left(\frac{1}{x}\right)}^4\times x$

• A. $(-25)$
• B. $(-27)$
• C. $(-15)$
• D. None of these

Answer 1

The correct option is B $(-27)$

We have

$x=(-3)$

We have to find the value of

${\left(x^2\right)}^3\times {\left(\frac{1}{x}\right)}^4\times x$

On putting the value of $x=-3$ we get,

$={\left\{{\left(-3\right)}^2\right\}}^3\times {\left(\frac{1}{-3}\right)}^4\times -3$

$={\left(-3\right)}^6\times {\left(\frac{1}{-3}\right)}^4\times -3$

$=-3\times -3\times -3\times -3\times -3\times -3\times \frac{1}{-3}\times \frac{1}{-3}\times \frac{1}{-3}\times \frac{1}{-3}\times -3$

$=\frac{3\times 3\times 3\times 3\times 3\times 3\times -3}{3\times 3\times 3\times 3}\left[weknowthat(-)\times (-)=(+)\right]$

$=3\times 3\times -3$

$=-27$

Hence the option (B) is true.