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STEP I 2004 question 1 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2004 question 1 solution
At this moment in the solution it is helpful that
Hence, "c" being "3" seems like a sensible choice:
Hence:
Hence:
Now consider that:
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![(c - d \sqrt{2})^{3} \\ = (c - d \sqrt{2})(c - d \sqrt{2})^{2} \\ = (c - d \sqrt{2})((c^{2} + 2d^{2}) - 2cd \sqrt{2}) \\ = \left[ c(c^{2} + 6d^{2}) \right] - \left[ d(3c^{2} + 2d^{2} \right] \sqrt{2}](http://thestudentroom.co.uk/../latexrender/pictures/ad55465c82eff1bb93f133276c5e3ff1.png "(c - d \sqrt{2})^{3} \\ = (c - d \sqrt{2})(c - d \sqrt{2})^{2} \\ = (c - d \sqrt{2})((c^{2} + 2d^{2}) - 2cd \sqrt{2}) \\ = \left[ c(c^{2} + 6d^{2}) \right] - \left[ d(3c^{2} + 2d^{2} \right] \sqrt{2}")
, as a solution can be spotted:
![3 - 2 \sqrt{2} = \sqrt[3]{ 99 - 70 \sqrt{2} }](http://thestudentroom.co.uk/../latexrender/pictures/586699169eed00f4abdce77ae98ac9d6.png "3 - 2 \sqrt{2} = \sqrt[3]{ 99 - 70 \sqrt{2} }")
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![(c + d \sqrt{2})^{3} = \left[ c(c^{2} + 6d^{2}) \right] + \left[ d(3c^{2} + 2d^{2} \right] \sqrt{2}](http://thestudentroom.co.uk/../latexrender/pictures/89abfbeb0423b6c655006d6c4a088c33.png "(c + d \sqrt{2})^{3} = \left[ c(c^{2} + 6d^{2}) \right] + \left[ d(3c^{2} + 2d^{2} \right] \sqrt{2}")
, hence:
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