Solve (y+1)^3=8

$$( y + 1 ) ^ { 3 } = 8$$

Solve for y

$$y=1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(y+1\right)^{3}.

$$y^{3}+3y^{2}+3y+1=8$$

Subtract 8 from both sides.

$$y^{3}+3y^{2}+3y+1-8=0$$

Subtract 8 from 1 to get -7.

$$y^{3}+3y^{2}+3y-7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±7,±1$$

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.

$$y=1$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}+3y^{2}+3y-7 by y-1 to get y^{2}+4y+7. Solve the equation where the result equals to 0.

$$y^{2}+4y+7=0$$

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 7 for c in the quadratic formula.

$$y=\frac{-4±\sqrt{4^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$y=\frac{-4±\sqrt{-12}}{2}$$

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.

$$y=1$$