nice exponent math simplification | find the value of x 

I reduced 87/21 to 3 from the beginning which reduced the working going forward

Nice exponent math simplification: (x⁷ + x⁵ + x³)/(x⁶ + x⁵ + x⁴) = 81/27; x = ? (x⁷ + x⁵ + x³)/(x⁶ + x⁵ + x⁴) = [(x³)(x⁴ + x² + 1)]/[(x⁴)(x² + x + 1)] = [3(27)]/27 = 3 x ≠ 0, x ≠ ± 1; (x⁴ + x² + 1)/(x² + x + 1) = [(x⁶ - 1)/(x² - 1)]/(x² + x + 1) = 3x [(x³)² - 1]/[(x + 1)(x - 1)(x² + x + 1)] = [(x³ + 1)(x³ - 1)]/[(x + 1)(x³ - 1)] = 3x (x³ + 1)/(x + 1) = [(x + 1)(x² - x + 1)]/(x + 1) = 3x, x² - x + 1 = 3x x² - 4x + 1 = 0, (x - 2)² = 3 = (√3)², x - 2 = ± √3; x = 2 ± √3 Answer check: (x⁷ + x⁵ + x³)/(x⁶ + x⁵ + x⁴) = 81/27; x² - x + 1 = 3x x = 2 ± √3: x² - 4x + 1 = 0, x² - x + 1 = 3x; Confirmed Final answer: x = 2 + √3 or x = 2 - √3

interesting 👏👏

Both numerator and denominator divide by x⁵: [x²+1+(1/x²)]/[x+1+(1/x)]=3 x²+(1/x²)+1=3[x+(1/x)+1] x²+(1/x²)-3[x+(1/x)]-2=0 [x²+(1/x²)+2]-3[x+(1/x)]-4=0 [x+(1/x)]²-3[x+(1/x)]-4=0 a quadratic equation in x+(1/x). The root is x+(1/x)=½[3±sqrt(25)] =½(3±5) x+(1/x)=4 --> x²-4x+1=0 (x-2)²=3 x=2±sqrt(3) x+(1/x)=-1 --> x²+x+1=0 x=½[-1±sqrt(-3)] =½[-1±isqrt(3)]

(x² + 1 + 1/x²)/(x + 1 + 1/x) = 3 x² + 1 + 1/x² = 3(x + 1 + 1/x) (x + 1/x)² - 3(x + 1/x) - 4 = 0 x + 1/x = u u² - 3u - 4 = 0 (u - 4)(u + 1) = 0 u - 4 = 0 => u = 4 x + 1/x = 4 x² - 4x + 1 = 0 x = (4 ± 2√3)/2 *x = 2 ± √3* u + 1 = 0 => u = -1 x + 1/x = -1 x² + x + 1 = 0 *x = (-1 ± i√3)/2*

It was unsmart to keep rewriting 81/27, when that's just 3

原式：（x^2+1+x^-2)/(x+1+x^-1)=3 x^2+1+x^-2=3(x+1+x^-1)

Did you know that 81/27=3?