Question

find the value of
$\sqrt[3]{.001} \times 10$
​

Answer 1

Its an infinite series, modifying it:

√(3+x)=x. Because x is recurring

Squaring

3+x=x^2

X^2-x-3=0

X=(-b+-√(b^2-4ac))/2a

=(1+-√(1–4*1*-3))/2

=(1+-√13)/2

As it is loop of square roots with no subtraction, (1-√13)/2 is out of question.

Number =(1+√13)/2

Hope that helps

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