Question

Show that:

ii) $\frac{\sqrt[3]{3-512}}{\sqrt[3]{3343}}=\sqrt[3]{3\frac{-512}{343}}$

Answer 1

Step: Show that $\frac{\sqrt[3]{3-512}}{\sqrt[3]{3343}}=\sqrt[3]{3\frac{-512}{343}}$

Given: $\frac{\sqrt[3]{3-512}}{\sqrt[3]{3343}}=\sqrt[3]{3\frac{-512}{343}}$

L.H.S.$=\frac{\sqrt[3]{3-512}}{\sqrt[3]{3343}}$

$=\frac{-\sqrt[3]{3512}}{\sqrt[3]{3343}}$ $[\therefore \sqrt[3]{3-a}=-\sqrt[3]{3a}]$

$\begin{array}{cc}\text{2}& \text{512}\\ \text{2}& \text{256}\\ \text{2}& \text{128}\\ \text{2}& \text{64}\\ \text{2}& \text{32}\\ \text{2}& \text{16}\\ \text{2}& \text{8}\\ \text{2}& \text{4}\\ \text{2}& \text{2}\\ & \text{1}\end{array}$

$\begin{array}{cc}\text{7}& \text{343}\\ \text{7}& \text{49}\\ \text{7}& \text{7}\\ & \text{1}\end{array}$



$=\frac{\sqrt[3]{32\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2}}{\sqrt[3]{37\times 7\times 7}}$

$=\frac{-(2\times 2\times 2)}{7}$

$=\frac{-8}{7}$

$=-\frac{8}{7}$

Now, R.H.S.$=\sqrt[3]{3\frac{-512}{343}}$

$=-\sqrt[3]{3\frac{512}{343}}$ $[\therefore \sqrt[3]{3-a}=-\sqrt[3]{3a}]$


$=-\sqrt[3]{3\frac{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2}{7\times 7\times 7}}$

$=-\sqrt[3]{3\frac{8}{7}\times \frac{8}{7}\times \frac{8}{7}}$

$=-\frac{8}{7}$

$\because L.H.S=R.H.S.$

$\therefore \frac{\sqrt[3]{3-512}}{\sqrt[3]{3343}}=\sqrt[3]{3\frac{-512}{343}}$

Hence proved.