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...
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
IMPOSSIBLE
No.
It is x = -5
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
IMPOSSIBLE
No.
y2=x3+3x2
It is x = -5
No, it is not.
(xn+2-1)/(x2-1)
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
One equation with two unknowns usually does not have a solution.
It has two complex roots.
(a) y = -3x + 1
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7