What is the answer to solvefor k,x=k^{-1/3} ?

The answer to solvefor k,x=k^{-1/3} is k= 1/(x^3) {x 0}

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solvefor k,x=k^{-1/3}

solve for

k, x = k^{- \frac{1}{3}}

k = \frac{1}{x ^{3}} {x < 0 or x > 0}

Solution steps

x = k^{- \frac{1}{3}}

Take both sides of the equation to the power of

- 3 : \frac{1}{x ^{3}} = k

\frac{1}{x ^{3}} = k

Switch sides

k = \frac{1}{x ^{3}}

Verify Solutions:

k = \frac{1}{x ^{3}} {x < 0 or x > 0}

The solution is

k = \frac{1}{x ^{3}} {x < 0 or x > 0}

- x^{2} + 2 x + 1 = x

5 x - 6 = 3 x + 4

3 = 3 x - 8 + 1

\frac{x}{x ^{2} + 1} = 0

What is the answer to solvefor k,x=k^{-1/3} ?
The answer to solvefor k,x=k^{-1/3} is k= 1/(x^3) {x<0\lor x>0}