Thread: Urgent help needed on equation!!!

Hey, i have to complete an AS level core one maths exam paper but i am completely stuck on the following question: Solve the equation x6+26x3-27=0

I have nooooo idea what to do! If someone could please answer the question with a full solution so i can understand it i would be most grateful!!!

Thank you very much!

Hello, LoveDeathCab!

Solve: .x^6 + 26x^3 - 27 = 0

Sure, you know what to do . . . try to factor it . . .


We have: .(x³ - 1)(x³ + 27) .= .0

Then solve the two equations:

. . x³ - 1 .= .0 . → . x³ = 1 . → . x = 1

. . x³ + 27 .= .0 . → . x³ = -27 . → . x = -3

You are a genius! Thank you sooooo much! It was doing my head in! I feel rather stupid now! Haha thanks!

Originally Posted by LoveDeathCab

Hey, i have to complete an AS level core one maths exam paper but i am completely stuck on the following question: Solve the equation x6+26x3-27=0

I have nooooo idea what to do! If someone could please answer the question with a full solution so i can understand it i would be most grateful!!!

Thank you very much!

Originally Posted by Soroban

Hello, LoveDeathCab!

Sure, you know what to do . . . try to factor it . . .


We have: .(x³ - 1)(x³ + 27) .= .0

Then solve the two equations:

. . x³ - 1 .= .0 . → . x³ = 1 . → . x = 1

. . x³ + 27 .= .0 . → . x³ = -27 . → . x = -3

I am surprised that Soroban did not mention this: you are missing 4 complex solutions here. (Perhaps the "AS level core exam" means something to him. It doesn't to me!) If you want them, this is how to get them:

x^6 + 26x^3 - 27 = 0

(x^3 - 1)(x^3 + 27) = 0

Now,
x^3 - 1 = (x - 1)(x^2 + x + 1)
x^3 + 27 = (x + 3)(x^2 - 3x + 9)

So:
x^6 + 26x^3 - 27 = (x^3 - 1)(x^3 + 27) = (x - 1)(x^2 + x + 1)(x + 3)(x^2 - 3x + 9) = 0

To get the solutions set each individual factor equal to 0.

I get
x = 3/2 (+/-) I*(3/2)*sqrt{3}, (-1/2) (+/-) I*(1/2)sqrt{3}, 1, -3
where I^2 = -1.

-Dan