What are the solutions to the equation r^3= 243/72 ?

The solutions to the equation r^3= 243/72 are r= 3/2 ,r=-3/4+(3sqrt(3))/4 i,r=-3/4-(3sqrt(3))/4 i

Find the zeros of r^3= 243/72

The zeros of r^3= 243/72 are r= 3/2 ,r=-3/4+(3sqrt(3))/4 i,r=-3/4-(3sqrt(3))/4 i

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r^{3} = \frac{243}{72}

r = \frac{3}{2}, r = - \frac{3}{4} + \frac{3 3}{4} i, r = - \frac{3}{4} - \frac{3 3}{4} i

Solution steps

r^{3} = \frac{243}{72}

Cancel the common factor:

9

r^{3} = \frac{27}{8}

For

x^{3} = f (a)

the solutions are

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

r = 3 \frac{27}{8}, r = 3 \frac{27}{8} \frac{- 1 + 3 i}{2}, r = 3 \frac{27}{8} \frac{- 1 - 3 i}{2}

3 \frac{27}{8} = \frac{3}{2}

Simplify

3 \frac{27}{8} \frac{- 1 + 3 i}{2} : - \frac{3}{4} + \frac{3 3}{4} i

Simplify

3 \frac{27}{8} \frac{- 1 - 3 i}{2} : - \frac{3}{4} - \frac{3 3}{4} i

r = \frac{3}{2}, r = - \frac{3}{4} + \frac{3 3}{4} i, r = - \frac{3}{4} - \frac{3 3}{4} i

y^{3} = y + 24

d^{4} - 4 d^{3} + 8 d^{2} - 8 d + 4 = 0

(100) = 2 x^{100} ln (100) + 20

x^{3} + x^{2} + 3 x = - 4

x^{8} + x^{4} + 1 = 0

What are the solutions to the equation r^3= 243/72 ?
The solutions to the equation r^3= 243/72 are r= 3/2 ,r=-3/4+(3sqrt(3))/4 i,r=-3/4-(3sqrt(3))/4 i
Find the zeros of r^3= 243/72
The zeros of r^3= 243/72 are r= 3/2 ,r=-3/4+(3sqrt(3))/4 i,r=-3/4-(3sqrt(3))/4 i