Solve (a-2)^3=8

$$( a - 2 ) ^ { 3 } = 8$$

Solve for a

$$a=4$$

Solution Steps

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

$$a^{3}-6a^{2}+12a-8=8$$

Subtract 8 from both sides.

$$a^{3}-6a^{2}+12a-8-8=0$$

Subtract 8 from -8 to get -16.

$$a^{3}-6a^{2}+12a-16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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.

$$a=4$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}-6a^{2}+12a-16 by a-4 to get a^{2}-2a+4. Solve the equation where the result equals to 0.

$$a^{2}-2a+4=0$$

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, -2 for b, and 4 for c in the quadratic formula.

$$a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$a=\frac{2±\sqrt{-12}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=4$$