In a two-dimensional world, the cone will look like an isosceles triangle.

We can cut this "isosceles" triangle into two equal right angled triangles, involving a height of x cm, a horizontal base of r cm and a hypotenuse of 10 cm.

Using the Pythagoras' / Pythagorean Theorem,
x² + r² = 10²
r² = 100 - x²

The volume of this cone
= 1/3 times π times radius² times height
= 1/3 times π times r² times x
= 1/3 times π times (100 - x²) times x
= 1/3 times π times x times (100 - x²)
= 1/3 times π times (100x - x³)
= -1/3 times π times x³ + 1/3 times π times 100x

Because the volume of the cone is 91π cm³,

1/3 times π times (100x - x³) = 91π
1/3 times (100x - x³) = 91
100x - x³ = 273
0 = x³ - 100x + 273
x³ - 100x + 273 = 0

Now, we express this cubic function as
x³ - 100x + 273 = (ax - 3) (x² + bx + c)

Clearly a = 1 because the coefficient of x³ on the left side of the equation is 1 (so we need to keep balance on the other side too!)

x³ - 100x + 273 = (x - 3) (x² + bx + c)

From here, you can do long division (if you know how), or you can compare coefficients like what I would do next.

Comparing constants (terms without x),
273 = (-3) (c)
c = -91

x³ - 100x + 273 = (x - 3) (x² + bx - 91)

Now, you can compare the coefficients of x² or to substitute a good value of x.

Letting x = 1,
1³ - 100 (1) + 273 = (1 - 3) [1² + b(1) - 91]
174 = (-2) (b - 90)
-87 = b - 90
90 - 87 = b
b = 3

x³ - 100x + 273 = (x - 3) (x² + 3x - 91)

When x³ - 100x + 273 = 0,
(x - 3) (x² + 3x - 91) = 0
x - 3 = 0 or x² + 3x - 91 = 0

You can continue from here on; the quadratic equation requires the use of a quadratic formula to obtain the value(s) of x.

Thank you so much this helped me a lot!