+499 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
