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In algebra, a cubic equation in one variable is an equation of the form + + + = in which a is not zero. The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation.
Casus irreducibilis (from Latin 'the irreducible case') is the name given by mathematicians of the 16th century to cubic equations that cannot be solved in terms of real radicals, that is to those equations such that the computation of the solutions cannot be reduced to the computation of square and cube roots.
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
The nested radicals in this solution cannot in general be simplified unless the cubic equation has at least one rational solution. Indeed, if the cubic has three irrational but real solutions, we have the casus irreducibilis, in which all three real solutions are written in terms of cube roots of complex numbers. On the other hand, consider the ...
The derivative of a cubic function is a quadratic function. A cubic function with real coefficients has either one or three real roots (which may not be distinct); [1] all odd-degree polynomials with real coefficients have at least one real root. The graph of a cubic function always has a single inflection point.
Descartes theory of geometric solution of equations uses a parabola to introduce cubic equations, in this way it is possible to set up an equation whose solution is a cube root of two. Note that the parabola itself is not constructible except by three dimensional methods.
One special working pup is getting his fifteen minutes of fame after helping a woman get home following a fun night out. On November 25, @aquariusgirly21, AKA D, shared a clip of the apartment ...
A counterexample by Ernst S. Selmer shows that the Hasse–Minkowski theorem cannot be extended to forms of degree 3: The cubic equation 3x 3 + 4y 3 + 5z 3 = 0 has a solution in real numbers, and in all p-adic fields, but it has no nontrivial solution in which x, y, and z are all rational numbers. [1]