For the expression [tex]x^3 y L + 4 L = 9 y_0[/tex], solve for [tex]x[/tex].

Possible answers are:
A.) [tex]\sqrt[5]{\frac{y_0 + 4}{L}}[/tex]
B.) [tex]\sqrt[3]{\frac{9 y_0 - 4 L}{y L}}[/tex]
C.) [tex]\sqrt{\frac{9 y_0 - 4 L}{y L}}[/tex]
D.) [tex]\sqrt[3]{\frac{81 - 16 L}{y}}[/tex]
E.) [tex](9 y_0 - 4 L)^3 \cdot y[/tex]



Answer :

To solve the equation [tex]\( x^3 y L + 4 L = 9 y_0 \)[/tex] for [tex]\( x \)[/tex], let's go step by step:

1. Isolate the [tex]\( x \)[/tex]-term: Move all terms except the term containing [tex]\( x \)[/tex] to one side of the equation.
[tex]\[ x^3 y L = 9 y_0 - 4 L \][/tex]

2. Solve for [tex]\( x \)[/tex]: We want to isolate [tex]\( x^3 \)[/tex], so divide both sides by [tex]\( y L \)[/tex].
[tex]\[ x^3 = \frac{9 y_0 - 4 L}{y L} \][/tex]

3. Take the cube root: To finally get [tex]\( x \)[/tex], we take the cube root of both sides of the equation.
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]

So, the solution to the equation [tex]\( x^3 y L + 4 L = 9 y_0 \)[/tex] is:
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]

The correct multiple-choice answer is:
[tex]\[ \boxed{\text{B}) \sqrt[3]{\frac{9 y 0 - 4 L}{y L}}} \][/tex]