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Find all solutions to $$7 \sqrt[3]{x} = x + 6$$

Apr 17, 2020

#1
+483
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If we cube both sides of this equation(I couldn't see an easy way out of this), we get:

$$343x = x^3 + 3x^2 * 6 + 3x * 6^2 + 6^3$$

$$343x = x^3 + 18x^2 + 108x + 216$$

Subtracting 343x on both sides, we get:

$$x^3 + 18x^2 -235x + 216 = 0$$

Realize that the sum of the coefficients is 0, which means that 1 is a root of this polynomial. Then we can factor out x-1 from this.

Doing polynomial division, we get this factored into:

$$(x-1)(x^2+19x-216)$$

We factor the right hand quadratic into:

$$x^2+19x-216 = (x-8)(x+27)$$

We have then factored our cubic into:

$$(x-1)(x-8)(x+27) = 0$$

This gives us our three solutions(by the fundamental theorem of algebra, there should be 3), as :

1, 8, and -27

Apr 17, 2020
#2
0

Let

$$\displaystyle x=y^{3},$$

then

$$\displaystyle 7y=y^{3}+6,\\ y^{3}-7y+6=0, \\ (y-1)(y^{2}+y-6)=0, \\ (y-1)(y-2)(y+3)=0,$$

etc. .

Guest Apr 18, 2020