Simplifying $ \;x({y^{3}}/{x^{4}})^{1/4}$

Question

I’m a little unsure how to simplify the following expression: $$ x\left(\frac{y^{3}}{x^{4}}\right)^{1/4} $$

According to the answer, this should get you $\;\; x y^{3/4} x^{-1} = y^{3/4} $.

My intuition tells me that when we bring up $ \,x^{4} \,$ from the denominator, we get $ \,x^{-4}\, $ (with a negative power). In general, I’m unsure how $ x $ on the outside should multiply with what is inside the brackets. (Following BEDMAS, brackets should go first, right?)

Answer 1

You have $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}$$ You can leave out the braces, if you apply the exponent $\frac14$ to each, nominator and denominator. Thus $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}=x\frac{y^\frac{3}{4}}{x^1}$$ Now as $\frac{1}{x} = x^{-1}$ you get $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}=x{y^\frac{3}{4}}{x^{-1}}=x^{1-1} y^\frac34= y^\frac34$$

Answer 2

$$ \begin{align} x\left(\frac{y^3}{x^4}\right)^{1/4} \\ \\ & = x(y^3x^{-4})^{1/4} \\ \\ & = xy^{3(1/4)}x^{-4(1/4)} \\ \\ & = xy^{3/4}x^{-1} \\ \\ & = xx^{-1}y^{3/4} \\ \\ & = \frac xx (y^{3/4}) \\ \\ & = y^{3/4} \\ \\ \end{align} $$