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Find the smallest numbers a, b, and c such that: a^2 = 2b^3 = 3c^5 What can you say about other solutions to this problem?

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

For the first part try squaring the whole expression.

For the second part, you could write $a = {\root 3\of {2+\sqrt{5}}}$ and $b= {\root 3\of {2-\sqrt{5}}}$, and use the identity $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab(a + b) + b^3.$$

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NRICH is part of the family of activities in the Millennium Mathematics Project.

NRICH is part of the family of activities in the Millennium Mathematics Project.