Question 1164447
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Let's look at a simpler, more familiar example, to see what the problem is with just taking the cube root on both sides.<br>
Suppose the equation is<br>
{{{x^2 = 4}}}<br>
If you just take the square root of both sides, you get x=2.  But you know there are two solutions, 2 and -2.<br>
Algebraically, to get all the solutions (in this case, both), you need to set everything equal to 0 and solve.<br>
{{{x^2-4=0}}}
{{{x-2)(x+2) = 0}}}<br>
x=2 OR x=-2.<br>
In that case, both solutions are real numbers.  In cases of equations with higher powers of x, that is often/usually not the case.  A polynomial equation of degree n has n roots; only in rare special cases are all the roots real.<br>
In your example, of the form x^3=A where A is any real number, there is always only one real root.<br>
So when you solve the equation by taking the cube root of both sides, you find only the real root; and you have lost the information you need to find the non-real roots.<br>
To find all three roots, you need to use the pattern for factoring the difference of cubes.  That will give you a linear factor which gives a real solution and a quadratic factor that gives two non-real solutions.<br>