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x(x^2+6x+18)(x+6)=409*23*17 On equating the above equation , we get x=17. Therefore , (17+4)^4=194481

we have x (x^2+6x+18)(x+6) = 159919

(x+3-3)(x^2+6x+18)(x+3+3) = 159919 [(x+3)^2 - 9] [(x+3)^2 + 9] = 159919 (x+3)^4 - 81 = 159919 (x+3)^4 = 160000 x+3 = 20 x = 17

therefore, (x+4)^4= (17+4)^4 = 21^4 = 194481

x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be

If we know x + 1/x =2 then what would be the value of x^2048 + 1/x^2048 +x^2047 - 1 / x^2047 + 1/x^2049 - x^2049 +2

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