Question

$\left(\frac{\sqrt{24}}{\sqrt{30}+\sqrt6}\right)^{24}=$ what?

Answer 1

$\left(\frac{\sqrt{24}}{\sqrt{30}+\sqrt6}\right)^{24}$

$=\left(\frac{\sqrt{6\times4}}{\sqrt{6\times5}+\sqrt6}\right)^{24}$

$=\left(\frac{\sqrt6\times\sqrt4}{\sqrt6\times\sqrt5+\sqrt6}\right)^{24}$

$=\left(\frac{\sqrt6\times2}{\sqrt6\left(\sqrt5+1\right)}\right)^{24}$

$=\left(\frac{\cancel{\sqrt6}\times2}{\cancel{\sqrt6}\left(\sqrt5+1\right)}\right)^{24}$

$=\left(\frac{2}{\sqrt5+1}\right)^{24}$

$=\left(\frac{2}{\sqrt5+1}\times\frac{\sqrt5-1}{\sqrt5-1}\right)^{24}$

$=\left(\frac{2\left(\sqrt5-1\right)}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\right)^{24}$

$=\left(\frac{2\left(\sqrt5-1\right)}{\left(\sqrt5\right)^2-1^2}\right)^{24}$

$=\left(\frac{2\left(\sqrt5-1\right)}{5-1}\right)^{24}$

$=\left(\frac{2\left(\sqrt5-1\right)}4\right)^{24}$

$=\left(\frac{\bcancel2\left(\sqrt5-1\right)}{\displaystyle\underset2{\bcancel4}}\right)^{24}$

$=\left(\frac{\left(\sqrt5-1\right)}2\right)^{24}$

$=\left(\frac{\left(\sqrt5-1\right)}2\right)^{3\times8}$

$=\left(\frac{\left(\sqrt5-1\right)^3}{2^3}\right)^8$

$=\left(\frac{\left(\sqrt5\right)^3-3\cdot\left(\sqrt5\right)^2\cdot1+3\cdot\sqrt5\cdot1^2-1^3}8\right)^8$

$=\left(\frac{5\sqrt5-3\cdot5\cdot1+3\cdot\sqrt5\cdot1-1}8\right)^8$

$=\left(\frac{5\sqrt5-15+3\sqrt5-1}8\right)^8$

$=\left(\frac{8\sqrt5-16}8\right)^8$

$=\left(\frac{8\left(\sqrt5-2\right)}8\right)^8$

$=\left(\frac{\bcancel8\left(\sqrt5-2\right)}{\bcancel8}\right)^8$

$=\left(\sqrt5-2\right)^8$

$\therefore \left(\frac{\sqrt{24}}{\sqrt{30}+\sqrt6}\right)^{24}=\left(\sqrt5-2\right)^8$ [EQ#1]

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Let, $x=\sqrt5-2$ [EQ#2]

Or, $x+2=\sqrt5$

Or, $\left(x+2\right)^2=\left(\sqrt5\right)^2$

Or, $x^2+2\cdot x\cdot2+2^2=5$

Or, $x^2+4x+4=5$

Or, $x^2+4x=5-4$

Or, $x^2+4x=1$

Or, $x^2=1-4x$ [EQ#3]

Or, $\left(x^2\right)^2=\left(1-4x\right)^2$

Or, $x^4=1^2-2\cdot1\cdot4x+\left(4x\right)^2$

Or, $x^4=1-8x+16x^2$

Or, $x^4=1-8x+16\left(1-4x\right)$ [useing EQ#3]

Or, $x^4=1-8x+16-64x$

Or, $x^4=17-72x$

Or, $\left(x^4\right)^2=\left(17-72x\right)^2$

Or, $x^8=\left(17\right)^2-2\cdot17\cdot72x+\left(72x\right)^2$

Or, $x^8=289-2448x+5184x^2$

Or, $x^8=289-2448x+5184\left(1-4x\right)$ [useing EQ#3]

Or, $x^8=289-2448x+5184-20736x$

Or, $x^8=5473-23184x$

Or, $\left(\sqrt5-2\right)^8=5473-23184\left(\sqrt5-2\right)$ [useing EQ#2]

Or, $\left(\sqrt5-2\right)^8=5473-23184\sqrt5+46368$

Or, $\left(\sqrt5-2\right)^8=51841-23184\sqrt5$

$\therefore \left(\frac{\sqrt{24}}{\sqrt{30}+\sqrt6}\right)^{24} = 51841-23184\sqrt5$ [useing EQ#1]