The cubic formula is rarely used and rarely talked about. This video explains how to depress a cubic polynomial into a form that works with the cubic formula. Watch Cube-root of Unity here: • Cube Root of Unity
It is possible to generalize this method for quartic (Euler did it) To solve depressed cubic you can substitute x = a+b but to solve depressed quartic substitution would be x = a+b+c Let see what we will get after this substitution x^4+px^2+qx+r x = a+b+c x^2 = a^2+b^2+c^2+2(ab+ac+bc) x^2 - (a^2+b^2+c^2) = 2(ab+ac+bc) (x^2 - (a^2+b^2+c^2))^2 = 4((a^2b^2+2a^2bc+a^2c^2)+2(ab+ac)bc + b^2c^2) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2 + 2(a^2bc + ab^2c + abc^2)) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2 + 2abc(a+b+c)) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2) + 8abc(a+b+c) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2) + 8abcx x^4 - 2(a^2+b^2+c^2)x^2 - 8abcx + (a^2+b^2+c^2)^2 - 4(a^2b^2 + a^2c^2 + b^2c^2) = 0 So after comparing coefficients we have following system of equations -2(a^2+b^2+c^2) = p -8abc = q (a^2+b^2+c^2)^2 - 4(a^2b^2 + a^2c^2 + b^2c^2) = r This system of equations can be easily transformed into Vieta formulas for cubic with the roots a^2 , b^2 , c^2
Although such generalization is possible i prefer to solve quartic by factoring into two quadratics For depressed quartic undetermined coefficients (x^2 - ax + b)(x^2 + ax + c) = x^4 + px^2 + qx + r works nicely but when given quartic is not already depressed i prefer to use differece of squares first
Your claim at 17:20 that the other two complex conjugate roots of your depressed cubic t³ + ²⁄₃t + ²⁰⁄₂₇ = 0 are obtained by multiplying the real root t = −²⁄₃ by ω = −¹⁄₂ + i·¹⁄₂√3 and by ω² = −¹⁄₂ − i·¹⁄₂√3 is incorrect, that is _not_ how it works. See my detailed comment on your previous video about solving depressed cubic equations. In fact, what you should do to obtain the other two roots is multiply one of the two cube roots by ω and the other by ω² _in either order._ So, since we have ³√(−10 + 6√3) = −1 + √3 ³√(−10 − 6√3) = −1 − √3 the roots of your cubic in t are t₁ = ¹⁄₃((−1 + √3) + (−1 − √3)) = ¹⁄₃(−2) = −²⁄₃ t₂ = ¹⁄₃((−¹⁄₂ + i·¹⁄₂√3)(−1 + √3) + (−¹⁄₂ − i·¹⁄₂√3)(−1 − √3)) = ¹⁄₃(1 + 3i) = ¹⁄₃ + i t₃ = ¹⁄₃((−¹⁄₂ − i·¹⁄₂√3)(−1 + √3) + (−¹⁄₂ + i·¹⁄₂√3)(−1 − √3)) = ¹⁄₃(1 − 3i) = ¹⁄₃ − i
I prefer x=a+b , I find it easier to memorize. (a+b)^3+(a+b)^2+(a+b)+1= a^3+a^2 * (3b+1)+a*(3b^2+2b+1)+(b3+b2+b+1) We need the coefficient of a^2 to be zero, which means (3b+1)=0 b=-1/3 (a + b)^3 + (a + b)^2 + a + b + 1 = a^3 + (2 a)/3 + 20/27
Great video brother but I want to point out that you can complete a cube it’s analogous to completing the square. This was done over 500 years ago. Scipione Del Ferro and others have done it.
hello first title sound intersting but maby the cubic needs mental help? xD 0:47 but if i can depress a cubic it has to be posible to get back the original form so in terms of the last video wouldn´t that be easyer ? 1:12 what is not posible?? how ever. 2:22 i think if you use the normal ruls of equantions it is unimportent wich number you choose it will work anyway! 3:19 ok i understand how it works but not whatfor you need it! 3:58 that way of using t i have seen it somewere befor and why you choose t and not z or r or what ever? 7:10 ok yes thats quite simple at this point 8:34 i think i schould learn the solution for (a+b)³ 9:53 thanks for simplify it for me! 10:35 wouldn´t it be easyer to write (t/3) ? 11:28 i think i would had simplifyed the sum in the brackets before instead of writing it out but allright! 11:51 ok yes now it´s fine! 13:51 yes i remeber that! but anyway less you if you wan´t to use that 15:55 ok much fun to calculate that with out an calculator xD 16:34 oh ok now it looks posible 17:34 allright 17:41 an could you please explain me why do you add quotes at the end? 17:42 and please also explain me the quote it self! yours sincerly K.Furry
Another RU-vidr shows a cubic formula derivation of one complex conjugate from a pair root found in ax^3 + bx^2 + cx + d = 0 whether or not it was a real root from the more than one real roots possibility. Because a cubic polynomial is at least from a real root (x + r) multiplied to a quadratic polynomial, the cubic root formula is related to if the resulting quadratic polynomial had real two roots or two imaginary numbers. The more versatile cubic formula is that of ax^3 + bc^2 + cx + d root formula.
Can this method be still used if all the roots are real? For example I have a cubic equation x³-10x²+31x-30=0 Then t³-7/3t-20/27=0 after this I am having trouble for computing t. Can anyone help me out?
if 6sqrt(3)-10 = (sqrt(a)-b)³ then you get 6sqrt(3) -10= (a+3b²)sqrt(a) - (3ab+b²). You see immediately that a=3 (sqrt!) and after solving (a+3b²)=6 you get b=1. So, (sqrt(a)-b)³ = (sqrt(3)-1)³ = 6sqrt(3) -10 and therfore sqrt(3)-1 = (6sqrt(3) -10)^(1/3) The solution of (-6sqrt(3) -10) yields -sqrt(3)-1. The sum (sqrt(3)-1) + (-sqrt(3)-1) = -2
Still I would like to see the whole process, anв where those formulas for 2 other roots emerge (I've seen at least 2 variantsб maybe more). And, perhaps I missed smth, but i didn't get how we see how many roots there are. Also I'm wondering how Cardano (or Tartaglia) found all the three solutions without using complex numbers.
I legitimately memorized both the cubic and quartic before but I think I memorized the quartic wrong bc the font of the thing I was looking at was too small
How to use cube root of unity to find all solutions ? ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lHe6iieqzBw.html Suppose omega is principal cube root of unity to get solutions you multiply one cubic radical by omega and the other one by omega^2 If you look at system which you get after solvig depressed cubic p = 3ab , -q = a^3 - b^3 it would be clear What about casus irreducibilis in my opinion it is worth considering this case because after using de Moivre theorem you will get trigonometric solution from complex cubic radicals
I commend you on not making a depressed cubic joke in this video. If humor is a low hanging fruit most people can't resist picking a fruit like that. I enjoyed the video, this was a new topic for me. Thank you!
One way I never see is to set x = (t-b)/3a, which also eliminates the quadratic term but scales the cubic to only have integers as coefficients and a monic leading coefficient. The numbers are bigger but working with integers is much nicer than working with fractions.
