well there's three answers to this question one of them is one and the other two are roots of a quadratic equation that include the imaginary number i.
forgot the exact answers but i remember that you have to factor x^3-1
One because if x3 = 1 then x must equal 1
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
x=4.
Subtract 1 from each side: x3 = 64; You might know that 4 cubed is 64, but if you don't then take log 64, divide it by 3 and take the antilog of your answer...
That's the same as x3 times x3 which equals x6
One because if x3 = 1 then x must equal 1
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
9
x=4.
2+4 (which equals 6) x3 (equals 18) +1 = 19
(xn+2-1)/(x2-1)
what x3 equals 63 is:21x3=63
Subtract 1 from each side: x3 = 64; You might know that 4 cubed is 64, but if you don't then take log 64, divide it by 3 and take the antilog of your answer...
x3 - 3x2 - 9x - 5 = 0 (x + 1)(x2 - 4x - 5) = 0 (x + 1)(x - 5)(x + 1) = 0 x ∈ {-1, 1, 5}
That's the same as x3 times x3 which equals x6
3xyz
The inverse of a number is 1 divided by that number. So the inverse of x3 + 1 is 1/(x3 + 1).