Find all the solutions of the equation x^6 + 63x^3 - 64 = 0?

Answer 1

This is a quadratic in x^3:
(x^3 - 1)(x^3 + 64) = 0
(x - 1)(x^2 + x + 1)(x + 4)(x^2 - 4x + 16) = 0
Roots are:
x = 1
x = ( -1 +/- i sqrt(3) ) / 2
x = -4
x = 2(1 +/- i sqrt(3) )

Answer 2

x^6 + 63x^3 - 64 = 0? Factor it

(x-1)(x+4)(x^2+x+1)(x^2-4x+16)

x-1 = 0
x = 1

x+ 4 = 0
x = -4

Use the Quadratic Formula to slove

x^2+x+1 = 0 False no sloution

x^2-4x+16 = 0 False no sloution

x = 1 or x =-4 are the solutions of the equation x^6 + 63x^3 - 64 = 0

Answer 3

the firs thing I would do is factor out a x^3

x^3((x^3)+63)-64=0
also lets put 64 over to the other side
x^3((x^3)+63)=64
now lets divide both sides by ((x^3)+63)
x^3=64/((x^3)+63)
now you can multiply the X^3 back to the other side so you have
x^6=1
and well 1 to anypower is 1 so
x=1

but you need to find all the solutions and there is more then simply x=1

lets go back to
x^3(x^3+63)=64
plugging in 1 into this equation will make them equal to each other, but we want to figure out another possible solution.

So we have x^3 rights and 64 should look pretty factorable. Hows about taking the cubed root of 64?
Looks like that ends up as 4.
so if x were 4 we would have 64(64+63)
that is way too big, but could we possibly make the inside a 1 or something?
Try -4

-4^3 = -64
-64(-64+63)
-64(-1)
64=64
so looks like we found our other solution

Answer 4

Let z=x^3
Then z^2-63z-64 = 0
(z-64)(z+1)=0
So z = -1 and z = 64
So x^3 =-1 --> x=-1, and x^3 = 64 --> x= 4
The other 4 roots are imaginary.