Solve for x cube root of 5x^2-61-4=0

$\sqrt[3]{5 x^{2} - 61} - 4 = 0$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{5 x^{2} - 61}$ as ${(5 x^{2} - 61)}^{\frac{1}{3}}$ .

${(5 x^{2} - 61)}^{\frac{1}{3}} - 4 = 0$

Add $4$ to both sides of the equation.

${(5 x^{2} - 61)}^{\frac{1}{3}} = 4$

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${({(5 x^{2} - 61)}^{\frac{1}{3}})}^{3} = 4^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(5 x^{2} - 61)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(5 x^{2} - 61)}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(5 x^{2} - 61)}^{\frac{1}{3} \cdot 3} = 4^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(5 x^{2} - 61)}^{\frac{1}{3} \cdot 3} = 4^{3}$

Rewrite the expression.

${(5 x^{2} - 61)}^{1} = 4^{3}$

Simplify.

$5 x^{2} - 61 = 4^{3}$

Simplify the right side.

Raise $4$ to the power of $3$ .

$5 x^{2} - 61 = 64$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Add $61$ to both sides of the equation.

$5 x^{2} = 64 + 61$

Add $64$ and $61$ .

$5 x^{2} = 125$

Divide each term in $5 x^{2} = 125$ by $5$ and simplify.

Divide each term in $5 x^{2} = 125$ by $5$ .

$\frac{5 x^{2}}{5} = \frac{125}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x^{2}}{5} = \frac{125}{5}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{125}{5}$

Simplify the right side.

Divide $125$ by $5$ .

$x^{2} = 25$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$

Solve for x cube root of 5x^2-61-4=0

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