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It depends.If you want three real numbers [math]x,y,z[/math] satisfying [math]x^3+y^3+z^3=33[/math] then there are infinitely many, and they are very easy to find.If you want three rational numbers [math]x,y,z[/math] satisfying [math]x^3+y^3+z^3=33[/math] then there are still infinitely many, though they are somewhat harder to find.If you want three integers, or whole numbers, satisfying that equation, then you'll need to be patient: nobody knows if such three numbers exist. [math]33[/math] is the smallest natural number for which it is unknown if it can be represented as a sum of three cubes of integers.UPDATE (March 2019): This was just solved!! Andrew Booker[1] just discovered how to represent [math]33[/math] as a sum of three integer cubes:[math]\displaystyle \begin{align} a &= 8866128975287528 \\ [/math][math]b[/math][math] &= -8778405442862239 \\ c &= -2736111468807040 \\ 33 &= a^3+b^3+c^3\end{align}[/math]For real solutions, consider[math]x=0, y=0, z=\sqrt[3]{33}[/math]or[math]x=y=z=\sqrt[3]{11}[/math]or[math]x=2, y=3, z=-\sqrt[3]{2}[/math].There clearly are infinitely many solutions: you can choose [math]x[/math] and [math]y[/math] to be whatever you want, and there's always exactly one [math]z[/math] that works, since every real number has exactly one real cube root.Want rational solutions? Here's a simple one:[math]\displaystyle (-4)^3+\left(-\frac{5}{3}\right)^3+\left(\frac{14}{3}\right)^3=33[/math].Once again, there are infinitely many such solutions, though that's not quite as obvious as the real case. If you pick [math]x[/math] and [math]y[/math] to be rational, there's generally no reason to assume that the corresponding [math]z[/math] will be rational as well – see the earlier examples of real solutions.But in fact, there are infinitely many ways of representing any rational number as the sum of three cubes of rationals. This was first discovered in 1825 by a schoolmaster from Leeds named S. Ryley, who published this discovery in “The Ladies’ Diary”, a publication which ran annually in London from 1704 to 1841, carrying the subtitle"Containing New Improvements in ARTS and SCIENCES, and many entertaining PARTICULARS: Designed for the USE AND DIVERSION OF THE FAIR SEX."One misses a time when the ladies of London were diverted by rational parametrizations of cubic equations.The problem of representing integers as sums of cubes of integers is very difficult.By simple considerations of residues modulo [math]9[/math], there are no such representations for numbers which leave a remainder of [math]4[/math] or [math]5[/math] upon division by [math]9[/math]. For all other numbers, it is believed that such representations exist, but they are chaotic in a rather… diverting manner.[math]1=1^3+0^3+0^3[/math][math]2=1^3+1^3+0^3[/math][math]3=1^3+1^3+1^3[/math][math]4[/math] is ruled out[math]5[/math] is ruled out[math]6=2^3+(-1)^3+(-1)^3[/math][math]7=2^3+(-1)^3+0^3[/math][math]8=2^3+0^3+0^3[/math][math]9=2^3+1^3+0^3[/math][math]10=2^3+1^3+1^3[/math][math]11=3^3+(-2)^3+(-2)^3[/math][math]12=10^3+7^3+(-11)^3[/math][math]13[/math] is ruled out[math]14[/math] is ruled out[math]15=2^3+2^3+(-1)^3[/math][math]16=2^3+2^3+0^3[/math][math]17=2^3+2^3+1^3[/math][math]18=3^3+(-2)^3+(-1)^3[/math][math]19=3^3+(-2)^3+0^3[/math][math]20=3^3+(-2)^3+1^3[/math][math]21=16^3+(-14)^3+(-11)^3[/math][math]22[/math] is ruled out[math]23[/math] is ruled out[math]24=2^3+2^3+2^3[/math][math]25=3^3+(-1)^3+(-1)^3[/math][math]26=3^3+(-1)^3+0^3[/math][math]27=3^3+0^3+0^3[/math][math]28=3^3+1^3+0^3[/math][math]29=3^3+1^3+1^3[/math][math]30=(-283059965)^3+(-2218888517)^3+2220422932^3[/math][math]31[/math] is ruled out[math]32[/math] is ruled out[math]33[/math]=??? Nobody knows. (EDIT: See above – now solved!)[math]34=3^3+2^3+(-1)^3[/math][math]35=3^3+2^3+0^3[/math][math]36=3^3+2^3+1^3[/math][math]37=4^3+(-3)^3+0^3[/math][math]38=4^3+(-3)^3+1^3[/math][math]39=134476^3+117367^3+(-159380)^3[/math][math]40[/math] is ruled out[math]41[/math] is ruled out[math]42=[/math]??? Nobody knows. (EDIT: in Sep 2019, this was solved as well!)The solutions given here are the smallest possible. The representation of [math]39[/math] was found in 1992 by Heath-Brown, Lioen and te Riele. The representation of [math]30[/math] was found in 1999 by Beck, Pine, Tarrant and Yarbrough Jensen[2].It is possible that some number not ruled out by the congruence condition has no such representation, though I feel it's unlikely. It is also possible that the numbers involved are beyond reach with present or even future technologies. Such is math: there are no guarantees.Footnotes[1] https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf[2] Mathematics of Computation

There is more than one way to interpret "solve" in the context of equations in general, and polynomial equations in particular. In practice, most often you would solve a cubic equation (3rd degree) by guessing one solution [math]x_1[/math], dividing the cubic by [math]x-x_1[/math] and then solving the resulting quadratic equation (2nd degree) as usual.You can also solve any such equation numerically, for example using Newton's method.However, when people talk about "solving an equation of the 3rd degree", they usually mean a solution by radicals. This means a way to write the three solutions as expressions involving the coefficients of the original equation, square roots, cubic roots, and arithmetic operations. The quadratic formula[math]x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/math]is, famously, a way to express the roots of the quadratic equation [math]ax^2+bx+c=0[/math] in this way.Why should anyone care?Honestly, cubic equations aren't that important. They do occur here and there "in real life" and in abstract math, but they're not really fundamentally vital. However, the method of solution has tremendous historical importance and considerable beauty, and therein lies its real value.It's amazing that mankind attempted to solve cubic equations for thousands of years, eventually succeeding after the dedicated efforts of countless people, yet so few today - even those with considerable mathematical training - are able to remember the algorithm well enough to be able to use it or explain it on demand, without looking it up. This was one of the key intellectual achievements of the renaissance, a significant breakaway from the frontiers of classical knowledge, and I think it deserves to be better known.It's not really that difficult, anyway. Ready?The Classical ApproachWe're trying to solve an equation that looks like [math]ax^3+bx^2+cx+d=0[/math] where the coefficients [math]a,b,c,d[/math] are some numbers. In fact, I'm going to focus on what seems like a very special case:(1) [math]x^3 + cx + d = [/math][math]0[/math][math] [/math].The reason for this focus is that reducing a general cubic to this form is fairly straightforward, so I'll leave this to the end. The real gem is this special case.You can play around with this equation for days and months to no avail - the algebraic manipulations we know and love just don't do anything here. The crucial idea is to perform a daring gambit, replacing the single unknown [math]x[/math] with a sum of two unknowns [math]p, q[/math]. So we just boldly write [math]x=p+q[/math].How can this be a good thing? We need to determine one number [math]x[/math], and now we have two unknowns to deal with? Well, sure, but we also have a new degree of freedom - we can impose another condition on [math]p[/math] and [math]q[/math] while allowing them to still add up to [math]x[/math]. Maybe that'll save us (hint: it does). This is the only part of the solution that you need to pull out of a hat. The rest is algebra.Let's put that [math]p+q[/math] in our equation to be solved (1). We get[math](p+q)^3 + c(p+q) + d = [/math][math]0[/math][math][/math]Expanding the cube, this is[math]p^3 + q^3 + 3p^2q + 3pq^2 + c(p+q) + d = [/math][math]0[/math][math][/math].We leave the cubes alone but observe that there's another [math]p+q[/math] hiding in the mixed terms:[math]p^3 + q^3 + 3pq(p+q) + c(p+q) + d = [/math][math]0[/math][math][/math].We see a repeated term [math]p+q[/math], so we naturally collect the terms that contain it:[math] p^3 + q^3 + (3pq+c)(p+q) + d = [/math][math]0[/math][math][/math].Now comes the kicker: remember that extra degree of freedom we have? We can use it to eliminate this whole mess in the middle. Quite simply, we nonchalantly impose that [math]3pq+c[/math] is to vanish.[math]3pq+c = [/math][math]0[/math][math][/math]Or:(2) [math] q = -\frac{c}{3p}[/math].Our equation is now:[math]p^3 - \frac{c^3}{27p^3} + d = [/math][math]0[/math][math][/math].And we're done!!Wait. Why are we done?Well, if you've solved many high-school quadratic exercises, you may recognize that the last equation is merely a quadratic in disguise. There's only one variable now, and that is [math]p[/math]. It only appears in the equation as [math]p^3[/math], so we might as well call [math]z = p^3[/math] and get[math]z - \frac{c^3}{27z} + d = [/math][math]0[/math][math][/math]or[math]z^2 + dz - \frac{c^3}{27}= [/math][math]0[/math][math][/math],A measly quadratic. Solve it, find the possible values of [math]z[/math], infer the possible values of [math]p[/math] (including the complex ones!), determine [math]q[/math] (remember equation (2)), and add them up to get [math]x[/math].TADAAA!!!A Worked ExampleLet's try to go through those steps with a concrete equation:[math]x^3 - 7x + [/math][math]6[/math][math] = [/math][math]0[/math][math] [/math].Once again, the way you really solve this is by guessing a solution (quite easy in this case), eliminating that solution and solving the quadratic. Takes about 2 minutes here. But no - we're going to follow the recipe. It's going to be way harder, introducing complex numbers along the way, but the beauty of it is that it's a recipe, requiring no guesses and guaranteed to work. Always.So, we make [math]x=p+q[/math],[math]p^3+q^3+3p^2q+3pq^2 - 7(p+q) + [/math][math]6[/math][math] = [/math][math]0[/math][math][/math][math]p^3 + q^3 + 3pq(p+q) - 7(p+q) + [/math][math]6[/math][math] = [/math][math]0[/math][math][/math][math]p^3 + q^3 + (3pq-7)(p+q) + [/math][math]6[/math][math] = [/math][math]0[/math][math][/math]We now declare that[math]3pq = 7[/math], or[math]q = \frac{7}{3p}[/math]getting[math]p^3 + \frac{343}{27p^3} + [/math][math]6[/math][math] = [/math][math]0[/math][math][/math]or, letting [math]z = p^3[/math],[math]z + \frac{343}{27z} + [/math][math]6[/math][math] = [/math][math]0[/math][math][/math][math]z^2 + 6z + \frac{343}{27} = [/math][math]0[/math][math][/math]This is somewhat hairy, but we plow through and find[math]z = [/math][math]-3[/math][math] \pm \frac{10\sqrt{3}}{9}i[/math].The next task is to find the [math]p[/math]s which correspond to each of those [math]z[/math]s. This is a matter of extracting cube roots of a complex number, a tricky but not insurmountable task. Let's find the cube roots of the one with the [math]+[/math] sign (it turns out that this is enough - repeating the task for the other value will yield the same roots again). They are:[math]p_1 = -\frac{3}{2} + \frac{\sqrt{3}}{6}i[/math][math]p_2 = \frac{1}{2} - \frac{5\sqrt{3}}{6} i[/math][math]p_3 = 1 + \frac{2\sqrt{3}}{3} i[/math]For each of these, we calculate [math]q = 7/(3p)[/math], add[math]p+q[/math] and find the values of [math]x[/math], which turn out to be...[math]x_1 = p_1 + \frac{7}{3p_1} = [/math][math]-3[/math][math][/math][math]x_2 = p_2 + \frac{7}{3p_2} = 1[/math][math]x_3 = p_3 + \frac{7}{3p_3} = 2[/math].Well. The roots turned out to be fairly simple numbers: -3, 1 and 2. That certainly could have been guessed. The method is impressively inefficient in this particular case, forcing us to work with these complex expressions only to get such simple answers eventually. But again: this is foolproof recipe that works for any cubic equation of the form [math]x^3+b x + c = [/math][math]0[/math][math][/math].The General CubicSo far we dealt with the special case [math]x^3+bx+c=0[/math]. In the general case, you start with[math]ax^3+bx^2+cx+d = [/math][math]0[/math][math][/math].First, divide through by [math]a[/math] to get rid of the leading coefficient (we assume that [math]a \neq [/math][math]0[/math][math][/math], otherwise it's not a cubic). You now have something like this (renaming the variables to keep things simple):[math]x^3+bx^2+cx+d = [/math][math]0[/math][math][/math].The coefficient [math]b[/math][math][/math] is minus the sum of the roots. This is just a special case of Vieta's formulas, but if you're not familiar with them you can easily see this by writing[math]x^3+bx^2+cx+d = (x-x_1)(x-x_2)(x-x_3)[/math], expanding and equating coefficients of [math]x^2[/math].This means that if we shift [math]x[/math] by [math]b/3[/math], the sum of the roots of the new equation will be 0. Indeed, let [math]y = x+b/3[/math] or [math]x = y-b/3[/math], substitute in the equation, and observe how [math]x^3[/math] contributes a factor of [math]-by^2[/math] and the [math]bx^2[/math] term contributes [math]by^2[/math] making the quadratic term disappear.This is much easier to do in specific instances than by following the formulas. If you don't remember what you need to shift the variable by, just write [math]x = y - t[/math], substitute and then find the value of [math]t[/math] that makes the square term vanish.Once you're done solving the formula for [math]y[/math], which is now in the correct form, just remember to subtract [math]b/3[/math] from each solution to get the values of the original [math]x[/math].That's it!

Simple! (x - 3)^{3} = 3^{3} \implies (x - 3)^{3} - 3^{3} = 0 Recall that : \alpha^{3} - \beta^{3} = (\alpha - \beta)(\alpha^{2} + \beta^{2} + \alpha\beta) \implies (x - 3 - 3)((x - 3)^{2} + 3^{2} + 3(x - 3)) = 0 \implies (x - 6)(x^{2} + 9 - 6x + 9 + 3x - 9) = 0 \implies x - 6 = 0 \implies \boxed{ x = 6 } \boxed { OR } \implies x^{2} - 3x + 9 = 0 Using the quadratic formula, x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a} Here a = 1, b = -3 and c = 9, \implies x = \dfrac{3 \pm \sqrt{(-3)^2 - 4(1)(9)}}{2(1)} \implies \boxed{ x = \dfrac{3 \pm 3\sqrt{3}\iota}{2}} Where, \iota = \sqrt{-1}. Therefore, \boxed{\boxed{ x = 6, \dfrac{3 + 3\sqrt{3}\iota}{2}, \dfrac{3 - 3\sqrt{3}\iota}{2}}} Thank you!(more)