algebra 2 help, algebra 2 help??????

x^3 + 64 = 0

x^3 = 0 - 64

x^3 = -64

cube root both expression to canel the power

x = -4

Use a^3 +b^3 = (a+b)(a^2 -ab+b^2)

So x^3 +64 =0

x^3 +4^3 = 0

(x+4)(x^2 -4x+16)=0

So x = -4 is the only real solution

x4 – 8x2 – 9 = 0

(x^2)^2 -8 (x^2) -9 = 0

(x^2 -9)(x^2 +1) = 0

(x-3)(x+3)(x^2 +1) =0

So x = 3 or -3

x^3 + 64 = 0

this is a 'sum of cubes'

a^3+b^3=(a+b)(a^2-ab+b^2)

x^3 + 64 = (x+4)(x^2-4x+16)

for sums and differences of cubes:

the sign in the first parenthesis mirrors the + for sums, or the - for differences

the first sign in the second parenthesis is oppostie

the second sign in the second parenthesis is always +

x^4-8x^2-9=0

substitute y for x^2

y^2-8y-9=0

(y-9)(y+1)

substitute x^2 back into the factors

(x^2-9)(x^2+1)

x^2-9 is a 'difference of squares'

(x+3)(x-3)(x^2+1)

x=i, +/-3

Do you want only real solutions?

x^3 + 64 = 0

(x + 4)(x^2 - 4x + 16) = 0

x + 4 = 0

x = -4

If you want complex solutions also, x^2 - 4x + 16 = 0

x^2 - 4x = -16

x^2 - 4x + 4 = -16 + 4

(x - 2)^2 = -12

x - 2 = ± 2i√3

x = 2 ± 2i√3, in addition to x = -4.

(x+4)(x^2-4x+16)=0

x+4=0

x=-4

x^2-4x+16=0

(4±√-48)/2

(4±4i√3)/2

2±2i√3

x={-4, 2+2i√3, 2-2i√3}

x4 – 8x2 – 9 = 0

(x^2-9)(x^2+1)=0

(x+3)(x-3)(x^2+1)=0

x=-3

x=3

x=±√-1

x={±i, -3, 3}

factor(x^3+64 = 0)

(x + 4)*(x^2 - 4*x + 16) = 0

x + 4 = 0

x1 = - 4

(x^2 - 4*x + 16) = 0

b^2 - 4*a*c = 16 - 4*1*16 = - 48 = - 16*3

x2 = ( - ( - 4 + sqrt( - 16*3))/2 = 2 + 2*i*sqrt(3)

x3 = 2 - 2*i*sqrt(3)

x^3 + 64 = 0

x^3 = -64

x^3 = - 4^3

x = 4 * sqrt(-1)

x = 4i

hope this helps!!!

Regards

SAM