What is x if x^3-1728=0?
We are given that x^3 - 1728 = 0 and we have to find x.
x^3 - 1728 = 0
=> x^3 = 1728
=> x^3 = 12^3
As the exponent is the same, we can equate the bases
This gives x = 12.
We notice that the number 1728 = 12^3.
We'll re-write the equation as a difference of squares:
x^3 - 12^3 = 0
We'll apply the formula:
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
x^3 - 12^3 = (x-12)(x^2 + 12x + 144)
If x^3 - 12^3 = 0 => (x-12)(x^2 + 12x + 144) = 0
We'll set each factor as zero:
x -12 = 0
x = 12
x^2 + 12x + 144 = 0
x1 = [-12+sqrt(144 - 576)]/2
Since sqrt-432 is not a real value, the equation has a single real solution.
The real solution of the equation is x = 12.