What are the solutions to the equation 1500=986(1+x)^3 ?

The solutions to the equation 1500=986(1+x)^3 are

Find the zeros of 1500=986(1+x)^3

The zeros of 1500=986(1+x)^3 are

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1500 = 986 (1 + x)^{3}

x = \frac{5 3 6}{3 493} - 1, x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} + i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}, x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} - i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}

Solution steps

1500 = 986 (1 + x)^{3}

Switch sides

986 (1 + x)^{3} = 1500

Divide both sides by

986

(1 + x)^{3} = \frac{750}{493}

For

g^{3} (x) = f (a)

the solutions are

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

Solve

1 + x = 3 \frac{750}{493} : x = \frac{5 3 6}{3 493} - 1

Solve

1 + x = 3 \frac{750}{493} \frac{- 1 + 3 i}{2} : x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} + i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}

Solve

1 + x = 3 \frac{750}{493} \frac{- 1 - 3 i}{2} : x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} - i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}

The solutions are

x = \frac{5 3 6}{3 493} - 1, x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} + i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}, x = \frac{49 3 ^{\frac{2}{3}} 3 2 ( - 2 ^{\frac{2}{3}} 3 493 - 5 3 3 )}{986} - i \frac{5 \cdot 3 ^{\frac{5}{6}} \cdot 49 3 ^{\frac{2}{3}} 3 2}{986}

x^{6} + 2 x^{3} + 2 = 0

x^{4} - 3 x^{3} - 5 x^{2} + 13 x + 6 = 0

28 x^{3} - 84 x^{2} = 0

x^{6} + 2 x^{3} - 3 = 0

4 x^{3} + 30 x^{2} + 70 x + 50 = 0

What are the solutions to the equation 1500=986(1+x)^3 ?
The solutions to the equation 1500=986(1+x)^3 are
Find the zeros of 1500=986(1+x)^3
The zeros of 1500=986(1+x)^3 are