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EXERCISE 4.2 · Q23

Q.Prove that (3+2)6+(3−2)6=970(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6 = 970.

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(a+b)6+(a−b)6=2[6C0a6+6C2a4b2+6C4a2b4+6C6b6](a+b)^6+(a-b)^6=2\left[{}^6C_0a^6+{}^6C_2a^4b^2+{}^6C_4a^2b^4+{}^6C_6b^6\right]. With a=3,b=2a=\sqrt3,b=\sqrt2 (a2=3,b2=2a^2=3,b^2=2): a6=27,a4=9,a2=3,b4=4,b6=8a^6=27,a^4=9,a^2=3,b^4=4,b^6=8. So $=2[1(27)+15(9)(2)+15(3)(4)+1(8)]=2[27+270 …

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