Top Answer

Realising that 125 is a perfect cube, this is actually pretty easy:

x3 + 125 = 0

x3 = -125

x = -5

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It is x = -5

x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)

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.

No.

It is a cubic polynomial in x and its value depends on the value of x.

cube root of 216 = 6 > X= 6

y2=x3+3x2

IMPOSSIBLE

(xn+2-1)/(x2-1)

x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)

No, it is not.

x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4

If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator

x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0

5

Since that isn't an equation, there is really nothing to solve.

Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.

No.

(x + 5)(x^2 - 5x + 25)

x3 + x2 - x = 0 ∴ x2 + x - 1 = 0 ∴ x2 + x + 0.25 = 1.25 ∴ (x + 0.5)2 = 1.25 ∴ x + 0.5 = ± 1.251/2 ∴ x = -0.5 ± 1.251/2

One equation with two unknowns usually does not have a solution.

(a) y = -3x + 1

2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7

It has two complex roots.

1-3x1=3x 1/3=x