Realising that 125 is a perfect cube, this is actually pretty easy:
x3 + 125 = 0
x3 = -125
x = -5
It is x = -5
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
IMPOSSIBLE
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
It is a cubic polynomial in x and its value depends on the value of x.
It is x = -5
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
IMPOSSIBLE
cube root of 216 = 6 > X= 6
x4 - 1.We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.
It is a cubic polynomial in x and its value depends on the value of x.
No.
y2=x3+3x2
No, it is not.
(xn+2-1)/(x2-1)
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4