Olympiad Mathematics | Learn how to solve the system for a and b quickly | Math Olympiad Training

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Olympiad Mathematics | Learn how to solve the system for a and b quickly | Math Olympiad Training

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[geoffreyparfitt]

The simplest algebra actually comes from substituting a = b + 8 which leads to a basic quadratic in two steps.

[mva]

I used a slightly different equation which look at the things directly, from the other end:

(a - b)^3 = a^3 - b^3 - 3ab(a - b)
8^3 = 344 - 3ab*8 ==> 512 = 344 - 24*ab ==> - 24ab = 168 ==> ab = -7.
The rest is the same, as in video. Keep up the great work.

[murdock]

Amazing task, excellent way of solution, many thanks! Another approach:
a – b = 8
a³ - b³ = 344 → a and b = integer (a or b has to be negative).
6³ = 216
7³ = 343
8³ = 512
→ a^3 - b^3 = 7^3 + 1 → a= 1; b = -7 🙂

[hansschotterradler]

I used the same difference of cubes equation as first step and substituted a-b with 8. Replace all "a" with 8+b and you end with a quadratic formula. Same result.

[guyhoghton]

Having got ab = -7 it is slightly quicker and neater to use Vieta's formula to solve for a and b:
p and q are the roots of x² - (p + q)x + pq = 0 (Vieta).
∴ a and (-b) are the roots of x² - [a + (-b)]x + a(-b) = 0
i.e. x² - (a - b)x - ab = 0
⇒ x² - 8x + 7 = 0
⇒ (x - 1)(x - 7) = 0
∴ a = 1, (-b) = 7 or a = 7, (-b) = 1
∴ (a, b) = (1, -7) or (7, -1).

[HappyFamilyOnline]

Very helpful video, 👍
thanks for sharing 😊

[defytyrantsofmississippi]

Good morning Professor how do you get B+ one times B +7 at time mark -2:18 from the above expression? Thank you and God bless you in advance

[rajeshvishwakarma]

Nice and fantastic 😊👌👍concept.
Great explanation.

Thanks🌹❤ pre math. 👍👌😊

[williamwingo]

Here's a more direct method:

a – b = 8; solve for b in terms of a:
b = a – 8;
a^3 – b^3 = 344; substitute b = a – 8:
a^3 – (a – 8)^3 = 344; expand (a – 8)^3 (I omit the gruesome details, but it was all done by hand, honest):
a^3 – (a^3 – 24a^2 – 192a – 512) = 344; remove the parentheses:
a^3 – a^3 + 24a^2 + 192a + 512 = 344; cancel out the a^3 terms and subtract 344 from both sides:
24 a^2 + 192a + 168 = 0; divide through by 24;
a^2 – 8a + 7 = 0; we could invoke the quadratic formula here, but this one is easy to factor:
(a – 1)(a – 7) = 0;
so a is either +1 or +7; and since b = a – 8,
this means b is either –7 or –1.
Check:
case 1: 1 – (–7) = 8;
1^3 – (–7)^3 = 1 – (–343) = 1 + 343 = 344;
case 2: 7 – (–1) = 8;
7^3 – (–1)^3 = 343 – (–1) = 343 + 1 = 344.
Thank you., ladies and gentlemen, I'm here all week. 🤠

[lovepreetsingh-rpzh]

iam jee advanced aspirant and i sove all his questions there are much helpful inn jeeadvanced exam. thanks sir

[hanswust]

Quite logic solution!

In hindsight, 8 = 7 + 1 and 344 = 7^3 + 1^3.

I'll try this approach next time.

[christianthomas]

as (a-b)^3 = a^3 - b^3 - 3b.a^2 + 3a.b^2, by substituting we get: 344 = 512 + 3b.a^2 + 3a.b^2. next after simplification we get: 56= ab(b-a). after substitution
for (b-a) we get: a.b=7. As a=b+8, we can substitute in a.b and get the quadratic: b^2 + 8.b+7=0 which gives two solutions: b=1 and b=-7 and follows: a=9 and a=1

[JPTaquari]

Not easy, but very nice problem!

a = 8 + b

(8+b)³ - b³ = 344

8³ + b³ + 3*8b ( 8 + b) = 344

24b² + 192b + 168 = zero ( divide by 24 )
b² + 8b + 7 = zero

Fazendo báskara, we have:
b' - 1
b" = - 7


Prova: 1 - 8 + 7 = zero
49 - 56 + 7 = zero


a-b =8
a - (-b) = 8
a = 1

a³ -b³ = 344
1 + 343 = 344


Bingo from Brazil

[bigm]

a and b fall by the wayside under Professor Premath’s numerical sword!

[studywithbhai]

Hi sir, your videos are really helpful

[MoonBillboardRileyReidMemes]

Hehe boy do I feel dumb. I haven't done math And I did science Olympiad with a wooden plane I didn't even build, this made me humble as hell.

Gotta study. Makes me feel good.

[mvs]

I figured both a and b can't be positive and b has to be negative which makes a and b must be <8. 7^3 is 343 so the answer is 7, -1 and 1, -7 also works. I know very little math but this just popped up in my head, lucky!

[БудівельникДОБРЯК]

Hello! Here's my solution:

We can represent difference between two cubes as next formula:
a³ – b³ = (a – b) • (a² + ab + b²);
We see a multiplier with its known value, we can get it from given system of equations (1st equation), thus:
(a – b) • (a² + ab + b²) = 8 • (a² + ab + b²);
Let's go back to last equation of the system:
8 • (a² + ab + b²) = 344;
a² + ab + b² = 43 – this will be our additional third equation;
Let's introduce the value of b in following way and paste it into 3rd equation:
a – b = 8 => a – 8 = b;
a² + ab + b² = a² + (a – 8) • a + (a – 8)² =
= a² + a² – 8a + a² – 16a + 64 =
= 3a² – 24a + 64 = 43;
3a² – 24a + 64 –43 = 0;
3a² – 24a + 21 = 0; / : 3
a² – 8a + 7 = 0;
a1 = 1; b1 = a1 – 8 = 1 – 8 = –7;
a2 = 7; b2 = a2 – 8 = 7 – 8 = –1.
Here are our solutions.
Answer: (1; –7) and (7; –1).

[jarikosonen]

344=8*(a^2+ab+b^2), then put here a=b+8 => 43=(b+8)^2+(b+8)*b+b^2 solve for b...?

[susennath]

I guess in 15 secs by trial method.

as 7^3=343.