Solve for x cube root of x^2-4+ cube root of 3x=0

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

Subtract $\sqrt[3]{3 x}$ from both sides of the equation.

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

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{x^{2} - 4}}^{3} = {(- \sqrt[3]{3 x})}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} - 4}$ as ${(x^{2} - 4)}^{\frac{1}{3}}$ .

${({(x^{2} - 4)}^{\frac{1}{3}})}^{3} = {(- \sqrt[3]{3 x})}^{3}$

Simplify the left side.

Simplify ${({(x^{2} - 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x^{2} - 4)}^{\frac{1}{3}})}^{3}$ .

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

${(x^{2} - 4)}^{\frac{1}{3} \cdot 3} = {(- \sqrt[3]{3 x})}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} - 4)}^{\frac{1}{3} \cdot 3} = {(- \sqrt[3]{3 x})}^{3}$

Rewrite the expression.

${(x^{2} - 4)}^{1} = {(- \sqrt[3]{3 x})}^{3}$

Simplify.

$x^{2} - 4 = {(- \sqrt[3]{3 x})}^{3}$

Simplify the right side.

Simplify ${(- \sqrt[3]{3 x})}^{3}$ .

Simplify the expression.

Apply the product rule to $- \sqrt[3]{3 x}$ .

$x^{2} - 4 = {(- 1)}^{3} {\sqrt[3]{3 x}}^{3}$

Raise $- 1$ to the power of $3$ .

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

Rewrite ${\sqrt[3]{3 x}}^{3}$ as $3 x$ .

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

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

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

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

Combine $\frac{1}{3}$ and $3$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Multiply $3$ by $- 1$ .

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

Solve for $x$ .

Add $3 x$ to both sides of the equation.

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

Factor $x^{2} - 4 + 3 x$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$(x - 1) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 4 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

Solve for x cube root of x^2-4+ cube root of 3x=0

Related Questions

Support LCMGCF.com by disabling your adblocker.