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Static Friction

evaluate (2 +...

Question

evaluate (2 +root(3))^7 +(2-root(3))^7

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Solution

$(2 + \sqrt{3})^{7} 7_{C_{0}} 2^{7} + 7_{C_{1}} 2^{6} \sqrt{3} + 7_{C_{2}} 2^{5} (\sqrt{3})^{2} + 7_{C_{3}} 2^{4} (\sqrt{3})^{3} + 7_{C_{4}} 2^{3} (\sqrt{3})^{4} + 7_{C_{5}} 2^{2} (\sqrt{3})^{5} + 7_{C_{6}} 2^{1} (\sqrt{3})^{6} + 7_{C_{7}} 2^{0} (\sqrt{3})^{7} (2 - \sqrt{3})^{7} = 7_{C_{0}} 2^{7} - 7_{C_{1}} 2^{6} \sqrt{3} + 7_{C_{2}} 2^{5} (\sqrt{3})^{2} - 7_{C_{3}} 2^{4} (\sqrt{3})^{3} + 7_{C_{4}} 2^{3} (\sqrt{3})^{4} - 7_{C_{5}} 2^{2} (\sqrt{3})^{5} + 7_{C_{6}} 2^{1} (\sqrt{3})^{6} - 7_{C_{7}} 2^{0} (\sqrt{3})^{7} (2 + \sqrt{3})^{7} S o, (2 + \sqrt{3})^{7} + (2 - \sqrt{3})^{7} = = 7_{C_{0}} 2^{7} + 7_{C_{1}} 2^{6} \sqrt{3} + 7_{C_{2}} 2^{5} (\sqrt{3})^{2} + 7_{C_{3}} 2^{4} (\sqrt{3})^{3} + 7_{C_{4}} 2^{3} (\sqrt{3})^{4} + 7_{C_{5}} 2^{2} (\sqrt{3})^{5} + 7_{C_{6}} 2^{1} (\sqrt{3})^{6} + 7_{C_{7}} 2^{0} (\sqrt{3})^{7} + 7_{C_{0}} 2^{7} - 7_{C_{1}} 2^{6} \sqrt{3} + 7_{C_{2}} 2^{5} (\sqrt{3})^{2} - 7_{C_{3}} 2^{4} (\sqrt{3})^{3} + 7_{C_{4}} 2^{3} (\sqrt{3})^{4} - 7_{C_{5}} 2^{2} (\sqrt{3})^{5} + 7_{C_{6}} 2^{1} (\sqrt{3})^{6} - 7_{C_{7}} 2^{0} (\sqrt{3})^{7} = 7_{C_{0}} 2^{7} + 7_{C_{2}} 2^{5} (\sqrt{3})^{2} + 7_{C_{4}} 2^{3} (\sqrt{3})^{4} + 7_{C_{6}} 2^{1} (\sqrt{3})^{6} = 2^{7} + \frac{7 \times 6}{2} 2^{5} \times 3 + \frac{7 \times 6 \times 5}{3 \times 2} 2^{3} 3^{2} + 7 \times 2 \times 3^{3} = 128 + 2016 + 2520 + 378 = 5042$

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