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Question Number 207226 by hardmath last updated on 10/May/24

$$\mathrm{Find}:\frac{{(2-\sqrt{2})}^8\cdot {(6+4\sqrt{2})}^8}{{(2+\sqrt{2})}^8}=?$$

Answered by Sutrisno last updated on 10/May/24

$${\left(\frac{(2-\sqrt{2})(6+4\sqrt{2})}{2+\sqrt{2}}\right)}^8$$$${(\frac{(2-\sqrt{2})(6+4\sqrt{2})}{2+\sqrt{2}}.\frac{2-\sqrt{2}}{2-\sqrt{2}})}^8$$$${\left(\frac{(4-4\sqrt{2}+2)(6+4\sqrt{2})}{4-2}\right)}^8$$$${\left(\frac{(6-4\sqrt{2})(6+4\sqrt{2})}{2}\right)}^8$$$${\left(\frac{36-32}{2}\right)}^8$$$${\left(\frac{4}{2}\right)}^8=256$$

Answered by Rasheed.Sindhi last updated on 10/May/24

$${\left(\frac{(2-\sqrt{2})\cdot (6+4\sqrt{2})}{(2+\sqrt{2})}\right)}^8$$$$={\left(\frac{(2-\sqrt{2})\cdot ({(2+\sqrt{2})}^2}{(2+\sqrt{2})}\right)}^8$$$$={\left((2-\sqrt{2})(2+\sqrt{2})\right)}^8$$$$={(4-2)}^8=256$$

Commented by hardmath last updated on 10/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{professors}$$

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