Solve 8y^3-27=0

$$8 y ^ { 3 } - 27 = 0$$

Solve for y

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$y=\frac{3}{2}$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 8y^{3}-27 by 2\left(y-\frac{3}{2}\right)=2y-3 to get 4y^{2}+6y+9. Solve the equation where the result equals to 0.

$$4y^{2}+6y+9=0$$

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 4 for a, 6 for b, and 9 for c in the quadratic formula.

$$y=\frac{-6±\sqrt{6^{2}-4\times 4\times 9}}{2\times 4}$$

Do the calculations.

$$y=\frac{-6±\sqrt{-108}}{8}$$

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.

$$y=\frac{3}{2}$$