Solve (1+1+alpha)^3=27

$$( 1 + 1 + \alpha ) ^ { 3 } = 27$$

Solve for α

$$\alpha =1$$

Solution Steps

Add 1 and 1 to get 2.

$$\left(2+\alpha \right)^{3}=27$$

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

$$8+12\alpha +6\alpha ^{2}+\alpha ^{3}=27$$

Subtract 27 from both sides.

$$8+12\alpha +6\alpha ^{2}+\alpha ^{3}-27=0$$

Subtract 27 from 8 to get -19.

$$-19+12\alpha +6\alpha ^{2}+\alpha ^{3}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\alpha ^{3}+6\alpha ^{2}+12\alpha -19=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -19 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±19,±1$$

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.

$$\alpha =1$$

By Factor theorem, \alpha -k is a factor of the polynomial for each root k. Divide \alpha ^{3}+6\alpha ^{2}+12\alpha -19 by \alpha -1 to get \alpha ^{2}+7\alpha +19. Solve the equation where the result equals to 0.

$$\alpha ^{2}+7\alpha +19=0$$

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, 7 for b, and 19 for c in the quadratic formula.

$$\alpha =\frac{-7±\sqrt{7^{2}-4\times 1\times 19}}{2}$$

Do the calculations.

$$\alpha =\frac{-7±\sqrt{-27}}{2}$$

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

$$\alpha \in \emptyset$$

List all found solutions.

$$\alpha =1$$