What are the solutions to the equation 18=b((2b)/3)^2 ?

The solutions to the equation 18=b((2b)/3)^2 are

Find the zeros of 18=b((2b)/3)^2

The zeros of 18=b((2b)/3)^2 are

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18 = b (\frac{2 b}{3})^{2}

Solution steps

18 = b (\frac{2 b}{3})^{2}

Switch sides

b (\frac{2 b}{3})^{2} = 18

Simplify

\frac{4 b ^{3}}{9} = 18

Multiply both sides by

9

\frac{9 \cdot 4 b ^{3}}{9} = 18 \cdot 9

4 b^{3} = 162

Divide both sides by

4

\frac{4 b ^{3}}{4} = \frac{162}{4}

b^{3} = \frac{81}{2}

For

x^{3} = f (a)

the solutions are

Simplify

2 k^{3} + 9 k^{2} + 13 k + 6 = 0

4 x^{3} + x^{2} + x - 48 = 0

7 y^{3} - 34 y^{2} - 5 y = 0

2 k^{3} - 3 k + 1 = 0

7 (7 x^{2})^{2} = 4 (7 x^{2}) + 1

What are the solutions to the equation 18=b((2b)/3)^2 ?
The solutions to the equation 18=b((2b)/3)^2 are
Find the zeros of 18=b((2b)/3)^2
The zeros of 18=b((2b)/3)^2 are