How do you solve $64x^3+27=0$ ?#?

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Answer 1 · 2015-09-29T23:24:51

Factor $64x^3+27 = (4x+3)(16x^2-12x+9)$

Hence $64x^3 + 27 = 0$ when $4x+3 = 0$ , that is when $x = -3/4$

In general $a^3+b^3 = (a+b)(a^2-ab+b^2)$

Putting $a=4x$ and $b=3$ we find:

$64x^3+27 = (4x)^3+3^3 = (4x + 3)((4x)^2 - (4x)*3 + 3^2)$

$= (4x+3)(16x^2 - 12x + 9)$

The remaining quadratic factor has no linear factors with Real coefficients.

You can tell that because its discriminant is negative:

$16x^2 - 12x + 9$ is of the form $ax^2+bx+c$ with $a = 16$ , $b = -12$ and $c = 9$ .

Its discriminant $Delta$ is given by the formula:

$Delta = b^2 - 4ac = (-12)^2 - (4xx16xx9) = 144 - 576$

$= -432$

In fact, $16x^2 - 12x + 9 = (4x+3omega)(4x+3omega^2)$

where $omega = -1/2 + sqrt(3)/2i$ is the primitive cube root of unity.

$omega$ is used when finding all the roots of a general cubic equation.

Answer 2 · 2015-09-30T01:07:19

$x= -3/4$

(As an alternate method)
$64x^3+27=0$

$rArr (4x)^3 = (-3)^3$

$rArr 4x=-3$

$rArr x= -3/4$

How do you solve 64x^3+27=064x^3+27=0?#?