Question

{256x^16/81y⁴}^-1/4

Please solve this ? Hardest question in rational number chapter

Answer 1

Answer:

$\left(\frac{256x^{16}}{81y^{4}}\right)^{\frac{-1}{4}}=\left(\frac{3y}{4x^{4}}\right)^{1}$

Step-by-step explanation:

$\left(\frac{256x^{16}}{81y^{4}}\right)^{\frac{-1}{4}}$

$=\left(\frac{4^{4}x^{16}}{3^{4}y^{4}}\right)^{\frac{-1}{4}}$

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

$=\left(\frac{4x^{4}}{3y}\right)^{-1}$

$=\left(\frac{3y}{4x^{4}}\right)^{1}$

Therefore,

$\left(\frac{256x^{16}}{81y^{4}}\right)^{\frac{-1}{4}}=\left(\frac{3y}{4x^{4}}\right)^{1}$

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