What are the solutions to the equation solvefor u,h(u)=(4u)^3 ?

The solutions to the equation solvefor u,h(u)=(4u)^3 are u=0,u=(sqrt(h))/8 ,u=-(sqrt(h))/8

Find the zeros of solvefor u,h(u)=(4u)^3

The zeros of solvefor u,h(u)=(4u)^3 are u=0,u=(sqrt(h))/8 ,u=-(sqrt(h))/8

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solve for

u, h (u) = (4 u)^{3}

u = 0, u = \frac{h}{8}, u = - \frac{h}{8}

Solution steps

h (u) = (4 u)^{3}

Expand

(4 u)^{3} : 64 u^{3}

u h = 64 u^{3}

Switch sides

64 u^{3} = u h

Move

u h

to the left side

64 u^{3} - u h = 0

Factor

64 u^{3} - u h : u (64 u^{2} - h)

u (64 u^{2} - h) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 64 u^{2} - h = 0

Solve

64 u^{2} - h = 0 : u = \frac{h}{8}, u = - \frac{h}{8}

The solutions are

u = 0, u = \frac{h}{8}, u = - \frac{h}{8}

2 x (2 x^{2} - 1) = 0

- \frac{x ^{3}}{18} - 3.6 x + 17.6 = 0

9 d x^{10} = 210

6 m^{3} - 12 m^{2} - 378 m = 0

(x - 2)^{3} + 1 = 0

What are the solutions to the equation solvefor u,h(u)=(4u)^3 ?
The solutions to the equation solvefor u,h(u)=(4u)^3 are u=0,u=(sqrt(h))/8 ,u=-(sqrt(h))/8
Find the zeros of solvefor u,h(u)=(4u)^3
The zeros of solvefor u,h(u)=(4u)^3 are u=0,u=(sqrt(h))/8 ,u=-(sqrt(h))/8