I have the radical $$\sqrt{1+\sqrt{5}(6-2\sqrt{5})^{1/4}}$$ for exam preparation (middle school):
I need to simplify it in natural numbers.
My attempt is:
We know the rule: $(a-b)^2=a^2-2ab+b^2$
Let's $a^2+b^2=6$, then $2ab=2\sqrt5$ and $ab=\sqrt5$, suppose that $b=1$, then $a=\sqrt5$.
$(6-2\sqrt{5})^{1/4}=(1^2-2\cdot1\cdot\sqrt{5}+(\sqrt{5})^2)^{1/4}=(\sqrt{5}-1)^{1/2}$
After first step:
$\sqrt{1+\sqrt{5}\sqrt{\sqrt{5}-1}}$
Question. What is a possible next step? Or possible error/typo in the task. The answer is 2.
As noted by Gerry Myerson, the expression is likely supposed to be $$\sqrt{(1+\sqrt5)\sqrt{6-2\sqrt5}}$$ $$= \sqrt{(\sqrt{5} + 1)(\sqrt{5} - 1)}$$ $$= \sqrt{5 - 1}$$ $$= 2$$
Remark: My intention to answer this question is less to answer the question (which I believe the OP is completely capable of doing themselves) and more to remove this question from the unanswered section.