How do you solve #x^ { 6} - 117x ^ { 3} = 1000#?

1 Answer
George C.
Apr 1, 2017

The real roots of the given equation are:

#x = 5" "# and #" "x = -2#

The other roots are:

#5omega#, #5omega^2#, #-2omega# and #-2omega^2#

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

Explanation:

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

We use both of these identities below, but first factor the given sextic equation as a quadratic in #x^3# as follows:

Note that #125*8=1000# and #125-8 = 117#

So we find:

#0 = x^6-117x^3-1000#

#color(white)(0) = (x^3-125)(x^3+8)#

#color(white)(0) = (x^3-5^3)(x^3+2^3)#

#color(white)(0) = (x-5)(x^2+5x+25)(x+2)(x^2-2x+4)#

Hence the real roots of the given equation are:

#x = 5" "# and #" "x = -2#

The remaining roots are complex, being:

#5omega#, #5omega^2#, #-2omega# and #-2omega^2#

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

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