Solve 5y^3=320

$$5 y ^ { 3 } = 320$$

Solve for y

$$y=4$$

Solution Steps

Divide both sides by 5.

$$y^{3}=\frac{320}{5}$$

Divide 320 by 5 to get 64.

$$y^{3}=64$$

Subtract 64 from both sides.

$$y^{3}-64=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±64,±32,±16,±8,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$y=4$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}-64 by y-4 to get y^{2}+4y+16. Solve the equation where the result equals to 0.

$$y^{2}+4y+16=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 4 for b, and 16 for c in the quadratic formula.

$$y=\frac{-4±\sqrt{4^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$y=\frac{-4±\sqrt{-48}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$y\in \emptyset$$

List all found solutions.

$$y=4$$