Thus is a good one. Not too easy and not too hard. Plus, whenever a difference of squares gives you a sum of cubes and a difference of cubes as its factors, there is a little symmetry on the page. Very nice yet again!
We put x= 6t then t^6-1= 0 (t^3-1)(t^3+1) = 0 (t-1)(t+1) (t^2+t+1)(t^2-t+1) = 0 t= 1, - 1 x = 6 - 6 ( real roots) for imaginary roots we will use quadratic formula on other two and then use x= 6t
More simple:
(x/2)^6=3^6, (x*1/2)^6=(1/2)^6 * (x)^6
(x)^6=[3^6/(1/2)^6]
(x)^6=[3^6 * 2^6]= (3*2)^6
x1,2=+-6.
Amazing
Thank you! Cheers! 😊
Thus is a good one. Not too easy and not too hard. Plus, whenever a difference of squares gives you a sum of cubes and a difference of cubes as its factors, there is a little symmetry on the page.
Very nice yet again!
Can i ask about the graphic tablet and the app which you use
We put x= 6t then
t^6-1= 0
(t^3-1)(t^3+1) = 0
(t-1)(t+1) (t^2+t+1)(t^2-t+1) = 0
t= 1, - 1
x = 6 - 6 ( real roots)
for imaginary roots we will use quadratic formula on other two and then use x= 6t
Excellent delivery! 👏
x_j=6[cos(jπ/3)+i*sin(jπ/3)],j=0,1,...,5 whee i=√(-1)
Math Olympiad Problem: (x/2)⁶ = 3⁶; x =?
First method:
(x/2)⁶ = 3⁶ = (6/2)⁶ = [(± 6)/2]⁶; x = ± 6, Missing four complex value roots, if available
Second method;
[(x/2)⁶](2⁶) = (3⁶)(2⁶), x⁶ = 6⁶, x⁶ - 6⁶ = (x³)² - (6³)² = (x³ - 6³)(x³ + 6³) = 0
(x - 6)(x² + 6x + 36)(x + 6)(x² - 6x + 36) = 0
x - 6 = 0, x = 6; x + 6 = 0, x = - 6; x² - 6x + 36 = 0 or x² + 6x + 36 = 0
(x - 3)² = - 27 = (3i√3)², x = 3 ± 3i√3 or (x + 3)² = (3i√3)², x = - 3 ± 3i√3
Answer check:
x = ± 6: (x/2)⁶ = (± 6/2)⁶ = 3⁶; Confirmed
x = 3 ± 3i√3 = 3(1 ± i√3): (x/2)² = [3(1 ± i√3)/2]² = (9/2)(- 1 ± i√3)
(x/2)³ = (x/2)²(x/2) = [(9/2)(- 1 ± i√3)][(3/2)(1 + i√3)] = (3³/4)(- 1 - 3) = - 3³
(x/2)⁶ = [(x/2)³]² = (- 3³)² = 3⁶; Confirmed
x = - 3 ± 3i√3 = 3(- 1 ± i√3): (x/2)² = [3(- 1 ± i√3)/2]² = (- 9/2)(1 ± i√3)
(x/2)³ = (x/2)²(x/2) = [(- 9/2)(1 ± i√3)][(3/2)(- 1 ± i√3)] = (- 3³/4)(- 1 - 3) = 3³
(x/2)⁶ = [(x/2)³]² = (3³)² = 3⁶; Confirmed
Final answer:
x = 6; x = - 6; Four complex value roots, if acceptable;
x = 3 + 3i√3; x = 3 - 3i√3; x = - 3 + 3i√3 or x = - 3 - 3i√3
This is dope! 👌
@@SpencersAcademy
Thanks 🙏
Me + or - 6😂
Thank you
You're welcome
Wow I wouldve thought since (x/2) = +- 3 --> x = 6, -6.
Bro why are you over compleicating
(x/2)⁶=3⁶
= ⁶√(x/2)⁶=⁶√3⁶
=x/2=3
=x=3×2
=X=6
Nice one! The degree of the equation is 6, so you need 5 more values of x. You've only got one.
(x/2)^6 = 3^6
or ((x/2)^6)^1/6 = (3^6)^2/6
or x/2 = 3
or x = 6 ans.
Please keep it simple.
Weldone! The degree of the equation is 6. So you need 4 more solutions of x.