Solving a simple cubic equation. A trick you should know!

This is a popular question on YouTube and someone requested that I present a solution. There is an important mathematical theorem you should know!

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[mahmoudaboualfa]

The easiest way to solve this problem is to factorize the variable part:
a³-a²=18
a²(a-1)=18
Now we look and think, what number raised to the power of 2 times another number gives us 18?
This lets us write the equation as:
a²(a-1)=9×2
By comparison:
a²=9 and a-1=2
a=±3 a=3
Since a has to be the same sign, we only have one answer a=3. (This step is faster for the brain to recognize than typing it.)
Then you go on by solving the rest by long division and get the complex answers to the quadratic equation.

[aguyontheinternet]

Right around here 1:50, I would recommend also bringing in the fact that it's impossible for any roots to be negative, as then the a^3, a^2, and the a^0 terms would all be negative, leaving no positives to balance it out to 0. Thus, no negative solutions exist, halving the guess and checking you'll need to do

[tse]

Observe that -18 can be rewritten as -27 + 9, s.t. difference of cubes and squares can be applied, we have
a^3 - 27 - (a^2 - 9) = 0
(a - 3)(a^2 + 3a + 9) - (a - 3)(a + 3) = 0
(a - 3)(a^2 + 2a +6) = 0
Thus, a = 3 or a = -1+-isqrt(5) .

[IAM-lwcj]

The number in the equation can be rewritten as 27-9 =3^3-3^2.
Now the equation can be written as:
a^3-3^3-(a^2-3^2)=0
Now we have a difference of two cubes and difference of two squares.
The problem now reduces to the same step after division.

[johncirillo]

Most Algebra textbooks will introduce synthetic division with the rational root theorem. Direct substitution, as you’ve demonstrated, is a little quicker. However, it’s nice to show your students the technique of synthetic division. Some students will gravitate to synthetic division over the long division technique and some will not. Nice video 😉👍

[thetroiimaster]

Use synthetic division...
You only use the constant, not variable. Make sure to use fillers, like +0a. Also bring the first constant down automatically.

Find the first value of a, which we discovered is 3(a-3=0), put that in top left for reference. Multiply the top left number by the furthest right number at the bottom, place it below the next term, and add. Put the answer in the bottom and repeat. See here:

3 | 1 -1 0 -18
— V 3
————————
1 2

3 | 1 -1 0 -18
— V 3 6 18
————————
1 2 6 0

After finishing, put the variables back, but with 1 less power, because we divided by a(what it equals)

1a^2+2a+6

Then solve from there. It even matches his, but is easier.

[donaldmcronald]

You can also use the Horner-Scheme to get the polynomial division. It‘s much faster and you just need the coefficients of your polynomial and fill in your guessed solution a=3.

[alexandra_reznikov]

This somehow reminds me of the time when one of my friends asked the teacher if 0°C + 0°C = 64°F. The teacher was so confused she didn't come to school after that for three days in a row.

Edit: I have summoned mathematicians and physicists in the comments section.

[bhimsharma]

That how SAT made simple looking equation to a compicated equation.

[Fyr]

At least for this example, you don't need to do long division once you have (a-3), you can simply compare coefficients. You can see from the fact its a cubic equation with coefficient of 1 for a^3, and by the fact that the constant term is 18, that it will be (a-3)(a^2 +ka + 6) Then simply compare coefficients and k must be 2.

Also alternatively to using the quadratic formula you can complete the square:
(a+1)^2 = -5
a+1 = +/- iroot5
a = -1 +/- iroot5

[NE]

3: 27 (which is 3 cubed) - 9 (which is 3 squared) = 18

I truly didn't even have to think about it. It just jumped out.

Thank you for your efforts. May you and yours stay well and prosper.

[ironfistgaming]

I am happy you discussed the general case also instead of focusing only on the current question

[mtaur]

When looking for whole number solutions, the form a^2*(a-1)=2*3*3 pretty quickly yields a=3, but the number of things to check grows for more highly composite numbers.

[nietzschessfan]

I do love this theorem... The first exam I won at University was about this topic... BUT when you can just guess and try small numbers into the equation, and it works, you can't tell with grandiosity it is a "Math Olimpiad Problem"

[santigamer]

You can use Ruffini as well, instead of the standard polynomial division

[wolfiegames]

instead of long division, you could instead use synthetic division to make it much faster.

[Unknown]

I did it by hit and trial.
a³-a²=18 (Given)

Trial 01:
When a=2, then a³-a²=(2)³-(2)²=8-4=4≠18
Trail 02:
When a=3, then a³-a²=(3)³-(3)²=27-9=18

Thus, a=3 satisfies the given equation.

[Lovuschka]

It is interesting to see that the difference between a² and a³ for 2 is 2*1*2, for 3 it is 3*2*3, for 4 it is 4*3*4, and so on. So it always is a²*(a-1).

[glennsampson]

Trial and error works. a=2 --> a^3 is too small. a=3 --> yep, that's the solution. That's a simple trick! 🙂
Also, a^2 is a factor of the l.h.s. So for an integer solution, we have to find a perfect square that is a factor of 18. There is only one a^2 = 9, thus a=3.

For additional roots, factor out (a-3) or use synthetic division, and solve the quadratic.

[Qruey]

This method reminds me of a rule i have been using since 2nd year of high school, called the Ruffini Rule. It works really similar to the method used here, but for less experienced people it may seem more viable in case they get messed up.

It works like this: After finding that 3 is an answer to the method, we put it in a table and rewrite the main equation, leaving the known term alone. Doing a set of calculations you end up with 0 at the end, and what you've written during those calculations are the coefficients to the equation you need to multiply the first term (a-3) with.