solving algebraic equations of degree greater than 2

Question

I observed that the roots of a quadratic equation can easily be found by multiplying the equation throughout by 4a (a being the coefficient of x squared) and "completing the square".So I tried to follow a similar method for the cubic but failed to find a corresponding multiplying factor to complete the cube. Is there one at all? What about equations of higher degrees?Would anyone kindly explain the theory behind this in detail?

Answer 1

While a general answer is provided by Galois theory, it's not especially difficult to see why computing the cube doesn't always work. First, we can informally observe that a quadratic can be written as $x^2 + a x + b$, which has two 'degrees of freedom'. The expression $(x-c)^2 + d$ that we aim to write it as also has two degrees of freedom, so it is not especially surprising that we can match coefficients. On the other hand, a cubic is written as $x^3 + a x^2 + b x + c$, and we aim to write it as $(x-d)^3 + e$. But there aren't enough degrees of freedom in this expression for it to be generally possible.

Let's look at a specific example now. Suppose we'd like to write $x^3 + 3x^2 + 2x + 1$ in the form $(x-d)^3 + e$, i.e. complete the cube. But $(x-d)^3 + e = x^3 - 3dx^2 + 3d^2 x - d^3 + e$, so it looks like we'd need to have $$-3d =3 \qquad 3d^2=2 \qquad -d^3+e=1$$ since the expressions need to be equal for all values of $x$. Looking at the first expression, $d=-1$. But this contradicts the second equation, because it would give $3 = 2$. For contrast with the quadratic case, in that case there would be only two equations to solve with two unknowns, which can be done. So, completing the cube is not in general possible.

Finally I'd recommend reading up on Cardano's method if you're interested in solving cubics, it's very possible to understand without any Galois theory.