# Find all the solutions of the equation 216x^3+2197=0?

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### 2 Answers

We have to solve the equation: 216x^3 + 2197 = 0

The highest power of x is 3, so we will have 3 roots.

216x^3 + 2197 = 0

=> 6^3*x^3 - 13^3 = 0

=> (6x + 13)(36*x^2 - 78*x + 169) = 0

6x + 13 = 0 => x1 = -13/6

36*x^2 - 78*x + 169 = 0

x2 = 78/ 72 + sqrt (78^2 - 4*169*36) / 72

=> x2 = 13/12 + i*sqrt ( 18252 )/72

=> x2 = 13/12 + i*sqrt ( 3*6*169 )/72

=> x2 = 13/12 + 6*13*i*sqrt 3 / 72

=> x2 = 13/12 + 13*i*sqrt 3 / 12

x3 = 13/12 - 13*i*sqrt 3 / 12

**The solutions of the equation are -13/6 , 13/12 + 13*i*sqrt 3 / 12 and 13/12 - 13*i*sqrt 3 / 12**

To solve the binomial equation, we'll apply the identity of the sum of cubes:

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

a^3 =216x^3

a = 6x

b^3 = 13^3

b = 13

216x^3+2197 = (6x+13)(36x^2 - 78x + 169)

If 216x^3+2197 = 0, then (6x+13)(36x^2 - 78x + 169) = 0

If a product is zero, then each factor could be zero.

6x+13 = 0

We'll subtract 13 both sides:

6x = -13

x1 = -13/6

36x^2 - 78x + 169 = 0

We'll apply the quadratic formula:

x2 = [78 + sqrt(6084-24336)]/72

x2 = (78+i*sqrt18252)/72

We'll factorize by 78 the numerator:

x2 = (78+78i*sqrt 3)/72

x2 = 13(1+isqrt3)/12

x3 = 13(1-isqrt3)/12

**The roots of the equation are: {-13/6 , 13(1+isqrt3)/12, 13(1-isqrt3)/12}.**